CP-4 Further algebra and functions — revision question pack

6 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section CP-4. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

CP-4.1 · Understand and use the relationship between roots and coefficients of polynomial equations up to quartic equations.

Explanation

For a monic cubic $x^3+ax^2+bx+c=0$ with roots $\alpha,\beta,\gamma$ , comparison with $(x-\alpha)(x-\beta)(x-\gamma)$ gives sum of roots $-a$ , sum of pair products $b$ and product $-c$ .
A monic quartic follows the alternating pattern: the elementary symmetric sums are $-a,b,-c,d$ .
Divide a non-monic equation by its leading coefficient first.
Further expressions are built from these sums: squares use $(\sum\alpha)^2$ , reciprocals use the ratio of the next-to-last symmetric sum to the product, and shifted products can be evaluated by substituting into the polynomial.
The roots need not be found explicitly.

Worked example

The roots of $x^3-6x^2+5x-2=0$ are $\alpha,\beta,\gamma$ . Find $\alpha^2+\beta^2+\gamma^2$ .

1. $\alpha+\beta+\gamma=6$ .
2. $\alpha\beta+\beta\gamma+\gamma\alpha=5$ .
3. $\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)$ .

Answer: $36-10=26$ .

Common mistakes

Don't fall into the trap of reading the sum of roots directly as the coefficient of $x^{n-1}$ without changing its sign.
Don't fall into the trap of using coefficient relations before dividing a non-monic polynomial by its leading coefficient.
Don't fall into the trap of replacing the sum of squares by the square of the sum and omitting the pair-product correction.

Exam tip

Write the required elementary symmetric sums from the coefficients before manipulating the target expression.

Tier 1 · Easy

1.
The roots of $x^3-5x^2+2x+8=0$ are $\alpha,\beta,\gamma$ . Without solving the equation, find $\alpha+\beta+\gamma$ , $\alpha\beta+\beta\gamma+\gamma\alpha$ and $\alpha\beta\gamma$ .
(3)
(Total for Question 1 is 3 marks)
2.
The roots of $x^3-4x^2-x+6=0$ are $\alpha,\beta,\gamma$ . Without finding the roots, find $\alpha^2\beta+\alpha\beta^2+\beta^2\gamma+\beta\gamma^2+\gamma^2\alpha+\gamma\alpha^2$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
The non-zero roots of $2x^4-3x^3-5x^2+7x-4=0$ are $\alpha,\beta,\gamma,\delta$ . Without solving the equation, find $\alpha+\beta+\gamma+\delta$ and $\frac1\alpha+\frac1\beta+\frac1\gamma+\frac1\delta$ .
(4)
(Total for Question 1 is 4 marks)
2.
The roots of $x^4-3x^3+ax^2+bx+6=0$ are $\alpha,\beta,\gamma,\delta$ . Given that $\alpha^2+\beta^2+\gamma^2+\delta^2=11$ and $\alpha^2\beta^2+\alpha^2\gamma^2+\alpha^2\delta^2+\beta^2\gamma^2+\beta^2\delta^2+\gamma^2\delta^2=25$ , determine $a$ and $b$ .
(5)
(Total for Question 2 is 5 marks)
3.
The roots of $f(x)=x^4-2x^3+3x^2+px+6$ are $\alpha,\beta,\gamma,\delta$ . Given that $(2-\alpha)(2-\beta)(2-\gamma)(2-\delta)=10$ , determine $p$ . Hence find $\displaystyle\sum_{i<j}\frac1{\alpha_i\alpha_j}$ , where $(\alpha_1,\alpha_2,\alpha_3,\alpha_4)=(\alpha,\beta,\gamma,\delta)$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
The roots of $x^3-4x^2+px-6=0$ are $\alpha,\beta,\gamma$ , and $\alpha^2+\beta^2+\gamma^2=10$ . Determine $p$ and then find $\alpha^3+\beta^3+\gamma^3$ without solving the cubic.
(5)
(Total for Question 1 is 5 marks)
2.
The roots of the equation $x^3-3x^2+px-2=0$ , where $p>0$ , are $\alpha,\beta,\gamma$ . Given that $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=13$ , determine $p$ and then find $\frac1{\alpha^2}+\frac1{\beta^2}+\frac1{\gamma^2}$ .
(6)
(Total for Question 2 is 6 marks)
3.
The roots of $x^4-4x^3+mx^2+nx+3=0$ are $\alpha,\beta,\gamma,\delta$ . Given that $\displaystyle\sum_{i<j}(r_i-r_j)^2=64$ , where $(r_1,r_2,r_3,r_4)=(\alpha,\beta,\gamma,\delta)$ , and that $\alpha\beta\gamma(\alpha+\beta+\gamma)+\alpha\beta\delta(\alpha+\beta+\delta)+\alpha\gamma\delta(\alpha+\gamma+\delta)+\beta\gamma\delta(\beta+\gamma+\delta)=-40$ , determine the value of $m$ and the value of $n$ . Hence find $(\alpha\beta\gamma)^2+(\alpha\beta\delta)^2+(\alpha\gamma\delta)^2+(\beta\gamma\delta)^2$ without finding the roots.
(7)
(Total for Question 3 is 7 marks)
4.
The roots of $x^4-6x^3+px^2+qx-180=0$ are $\alpha,\beta,\gamma,\delta$ . Given that $\alpha^2+\beta^2+\gamma^2+\delta^2=74$ and that $-5$ is one of the roots, determine the values of $p$ and $q$ . Hence factorise the polynomial completely and state the other three roots.
(7)
(Total for Question 4 is 7 marks)
5.
The roots of $x^3-5x^2+7x-3=0$ are $\alpha,\beta,\gamma$ . Form a polynomial equation whose roots are $\alpha\beta$ , $\beta\gamma$ and $\gamma\alpha$ . Hence find $\dfrac1{\alpha\beta-2}+\dfrac1{\beta\gamma-2}+\dfrac1{\gamma\alpha-2}$ .
(7)
(Total for Question 5 is 7 marks)

CP-4.2 · Form a polynomial equation whose roots are a linear transformation of the roots of a given polynomial equation (of at least cubic degree).

Explanation

If $f(t)=0$ has roots $\alpha_i$ and the required new roots are $x=a\alpha_i+b$ , with $a\ne0$ , rearrange the transformation to express the old root as $t=(x-b)/a$ , then substitute this into $f(t)=0$ .
Multiplying by a suitable power of $a$ clears denominators without changing the roots, after which the polynomial is expanded and collected in powers of $x$ .
For a shift, reverse the shift in the substitution; for a scaling, divide the new variable by the scale factor.
The final equation may be multiplied by any non-zero constant, but its degree and number of transformed roots must remain unchanged.

Worked example

The roots of $t^3+2t-4=0$ are $\alpha,\beta,\gamma$ . Form an equation with roots $2\alpha+3$ , $2\beta+3$ , $2\gamma+3$ .

1.Let $x=2t+3$ , so $t=(x-3)/2$ .
2.Substitute: $((x-3)/2)^3+2((x-3)/2)-4=0$ .
3.Multiply by $8$ : $(x-3)^3+8(x-3)-32=0$ .
4.Expand and collect terms.

Answer: $x^3-9x^2+35x-83=0$ .

Common mistakes

Don't fall into the trap of substituting $t=ax+b$ instead of rearranging the new-root transformation for the old root.
Don't fall into the trap of clearing a cubic denominator by $a$ rather than by $a^3$ .
Don't fall into the trap of changing the polynomial degree while expanding the transformed equation.

Exam tip

Define the new root, rearrange for the old variable, and display that substitution before expanding.

Tier 1 · Easy

1.
The roots of $t^3-3t+1=0$ are $\alpha,\beta,\gamma$ . Form a polynomial equation whose roots are $\alpha+2,\beta+2,\gamma+2$ .
(3)
(Total for Question 1 is 3 marks)
2.
The roots of $w^3+3w^2-4w-12=0$ are $2\alpha-1$ , $2\beta-1$ and $2\gamma-1$ . Form a polynomial equation whose roots are $\alpha,\beta,\gamma$ , giving your answer in the form $dx^3+ex+f=0$ , where $d$ , $e$ and $f$ are integers with no common factor and $d>0$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
The roots of $2t^3-t^2+4t-3=0$ are $\alpha,\beta,\gamma$ . Form a polynomial equation whose roots are $3\alpha-1$ , $3\beta-1$ and $3\gamma-1$ .
(4)
(Total for Question 1 is 4 marks)
2.
The roots of $t^3-2t^2+5t-1=0$ are $\alpha,\beta,\gamma$ . (a) Show that the equation with roots $2\alpha+3$ , $2\beta+3$ , $2\gamma+3$ is $u^3+du^2+eu+f=0$ , where $d$ , $e$ and $f$ are integers to be found. (3) (b) Hence find the equation with roots $1-(2\alpha+3)$ , $1-(2\beta+3)$ , $1-(2\gamma+3)$ , giving your answer in the form $x^3+ax^2+bx+c=0$ . (2)
(5)
(Total for Question 2 is 5 marks)
3.
The roots of $t^3-5t^2+2t+7=0$ are $\alpha,\beta,\gamma$ . Form a polynomial equation whose roots are $\alpha+2\beta+2\gamma$ , $2\alpha+\beta+2\gamma$ and $2\alpha+2\beta+\gamma$ , giving your answer in the form $x^3+px^2+qx+r=0$ , with integer coefficients $p$ , $q$ and $r$ to be found.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
The roots of $t^4-2t^3+t^2+3t-1=0$ are $\alpha,\beta,\gamma,\delta$ . Form a polynomial equation whose roots are $2-3\alpha$ , $2-3\beta$ , $2-3\gamma$ and $2-3\delta$ .
(6)
(Total for Question 1 is 6 marks)
2.
The roots of $t^3-3t^2+2t-5=0$ are $\alpha,\beta,\gamma$ . New roots are $m\alpha+n$ , $m\beta+n$ , $m\gamma+n$ , where $m>0$ . The equation with these roots is $x^3+3x^2+cx+d=0$ , where $c=-1$ . (a) Determine the values of $m$ and $n$ . (5) (b) Determine the value of $d$ . (2)
(7)
(Total for Question 2 is 7 marks)
3.
The roots of $2t^4-8t^3+3t^2+5t-7=0$ are $\alpha,\beta,\gamma,\delta$ . A constant $h$ is chosen so that the roots $\alpha+h$ , $\beta+h$ , $\gamma+h$ , $\delta+h$ have sum zero. (a) Determine $h$ and form a polynomial equation whose roots are these four shifted roots, giving your answer in the form $au^4+bu^3+cu^2+du+e=0$ , where $a$ , $b$ , $c$ , $d$ and $e$ are integers with no common factor and $a>0$ . (5) (b) Hence form a polynomial equation whose roots are $2(\alpha+h)-5$ , $2(\beta+h)-5$ , $2(\gamma+h)-5$ and $2(\delta+h)-5$ , giving your answer in the form $ax^4+bx^3+cx^2+dx+e=0$ , where $a$ , $b$ , $c$ , $d$ and $e$ are integers with no common factor and $a>0$ . (3)
(8)
(Total for Question 3 is 8 marks)
4.
The roots of $2t^4-3t^3+5t^2-4t+7=0$ are $3\alpha+1$ , $3\beta+1$ , $3\gamma+1$ and $3\delta+1$ . Form a polynomial equation whose roots are $2\alpha-5$ , $2\beta-5$ , $2\gamma-5$ and $2\delta-5$ , giving your answer in the form $au^4+bu^3+cu^2+du+e=0$ , where $a,b,c,d,e$ are integers with no common factor and $a>0$ . Hence find the sum of the four new roots.
(8)
(Total for Question 4 is 8 marks)
5.
The roots of $t^3-4t^2+t+2=0$ are $\alpha,\beta,\gamma$ . Form a polynomial equation whose roots are $2\alpha-\beta-\gamma$ , $2\beta-\gamma-\alpha$ and $2\gamma-\alpha-\beta$ . Hence find the sum of the squares of the new roots and the value of $(u_1+2)(u_2+2)(u_3+2)$ , where $u_1$ , $u_2$ and $u_3$ are the new roots.
(8)
(Total for Question 5 is 8 marks)

CP-4.3 · Understand and use formulae for the sums of integers, squares and cubes and use these to sum other series.

Explanation

The standard results are $\sum_{r=1}^n r=\dfrac{n(n+1)}2$ , $\sum_{r=1}^n r^2=\dfrac{n(n+1)(2n+1)}6$ and $\sum_{r=1}^n r^3=\left[\dfrac{n(n+1)}2\right]^2$ .
To sum a polynomial expression in $r$ , expand it, split the summation term by term and substitute the appropriate formulae.
A constant term $c$ contributes $cn$ when there are $n$ terms.
Different lower or upper limits may require subtracting an initial partial sum.
Factor before expanding the final expression where possible, and verify a closed form with a small value such as $n=1$ to expose limit or constant-term errors.

Worked example

Find a closed form for $\sum_{r=1}^n(2r^2+3r-1)$ .

1.Split the sum as $2\sum r^2+3\sum r-\sum1$ .
2.Substitute the standard results to obtain $\dfrac{n(n+1)(2n+1)}3+\dfrac{3n(n+1)}2-n$ .
3.Use a common denominator and simplify.

Answer: $\dfrac{n(4n^2+15n+5)}6$ .

Common mistakes

Don't fall into the trap of treating $\sum_{r=1}^n1$ as $1$ instead of $n$ .
Don't fall into the trap of using the square-sum formula for a cubic term.
Don't fall into the trap of changing the summation limits when splitting one sum into several sums.

Exam tip

Show the split into standard sums before substitution; this makes the method visible even if later algebra slips.

Tier 1 · Easy

1.
Evaluate $\sum_{r=1}^{20}r(r+1)$ .
(3)
(Total for Question 1 is 3 marks)
2.
Evaluate $\displaystyle\sum_{r=3}^{10}(r^2-r)$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Show that $\sum_{r=1}^{n}(3r^2-2r+4)=\frac{n(2n^2+n+7)}2$ .
(4)
(Total for Question 1 is 4 marks)
2.
You may assume the standard results for $\displaystyle\sum r$ and $\displaystyle\sum r^2$ . A constant $k$ satisfies $\displaystyle\sum_{r=1}^{8}(r^2+kr)=420$ . Determine $k$ .
(4)
(Total for Question 2 is 4 marks)
3.
You may assume the standard results for $\sum r$ and $\sum r^2$ . A positive integer $n$ satisfies $\displaystyle\sum_{r=1}^{n}(3r^2-r)=576$ . Determine $n$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Prove that $\sum_{r=1}^{n}r(r+1)(2r+1)=\frac{n(n+1)^2(n+2)}2$ . Hence evaluate the sum when $n=15$ .
(6)
(Total for Question 1 is 6 marks)
2.
(a) Show that $\displaystyle\sum_{r=1}^{k}(5r^2-3r+2)=\frac{k(5k^2+3k+4)}{3}$ . (4) (b) Hence show that $\displaystyle\sum_{k=1}^{n}\frac{k(5k^2+3k+4)}{3}=\frac{n(n+1)(5n^2+9n+10)}{12}$ . (4) (c) Hence find $\displaystyle\sum_{k=11}^{25}\frac{k(5k^2+3k+4)}{3}$ . (2)
(10)
(Total for Question 2 is 10 marks)
3.
You may assume the standard results for $\sum r$ and $\sum r^2$ . Let $S_n=\displaystyle\sum_{r=1}^{n}(ar^2+br+c)$ , where $a$ , $b$ and $c$ are constants. Given that $S_5=245$ , $S_8=924$ and the eighth term is $294$ , determine $a$ , $b$ and $c$ . Hence find $\displaystyle\sum_{r=7}^{18}(ar^2+br+c)$ .
(10)
(Total for Question 3 is 10 marks)
4.
(a) Show that $\displaystyle\sum_{r=1}^{n-1}r(n-r)=\dfrac{n(n-1)(n+1)}6$ . (4) (b) Hence show that $\displaystyle T_N=\sum_{n=2}^{N}\sum_{r=1}^{n-1}r(n-r)=\dfrac{N(N-1)(N+1)(N+2)}{24}$ . (4) (c) Find $T_{12}$ . (2)
(10)
(Total for Question 4 is 10 marks)
5.
(a) Show that $\displaystyle\sum_{r=1}^{n}(3r+2)^2=\frac{n(6n^2+21n+23)}2$ . (4) (b) The first $n$ terms of an arithmetic sequence are $5,8,11,\ldots,3n+2$ . Show that their mean is $\dfrac{3n+7}{2}$ and hence find the sum of the squared deviations of the terms from their mean. (5) (c) Find this sum for the first $12$ terms. (1)
(10)
(Total for Question 5 is 10 marks)

CP-4.4 · Understand and use the method of differences for summation of series including use of partial fractions.

Explanation

The method of differences rewrites the general term as a difference such as $f(r)-f(r+1)$ or $f(r)-f(r+k)$ . When consecutive terms are written out, most contributions cancel and only boundary terms remain.
Partial fractions often reveal this structure; the coefficients and index shift must be found exactly.
Write enough terms at the beginning and end of the finite sum to show the cancellation pattern, especially when the shift is greater than one.
The surviving first and last terms give the closed form.
For an infinite sum, first obtain the finite partial sum and then take its limit; cancellation is not a substitute for checking convergence.

Worked example

Find $\sum_{r=1}^n\dfrac1{(2r-1)(2r+1)}$ by differences.

1. $\dfrac1{(2r-1)(2r+1)}=\dfrac12\left(\dfrac1{2r-1}-\dfrac1{2r+1}\right)$ .
2.Writing the terms gives $\dfrac12[(1-1/3)+(1/3-1/5)+\cdots]$ .
3.All intermediate fractions cancel, leaving $\dfrac12(1-1/(2n+1))$ .

Answer: $\dfrac{n}{2n+1}$ .

Common mistakes

Don't fall into the trap of omitting the factor $1/2$ from the partial-fraction decomposition.
Don't fall into the trap of cancelling a boundary term that has no matching term of opposite sign.
Don't fall into the trap of taking an infinite-series limit before deriving the finite partial sum.

Exam tip

For 'use the method of differences', display the first two and final two terms before stating what cancels.

Tier 1 · Easy

1.
Use the method of differences to find $\sum_{r=1}^{n}\frac1{(r+2)(r+3)}$ .
(3)
(Total for Question 1 is 3 marks)

Tier 2 · Standard

1.
Find a closed form for $\sum_{r=1}^{n}\frac1{r(r+2)}$ . Hence evaluate the sum for $n=10$ .
(6)
(Total for Question 1 is 6 marks)
2.
(a) Show that $\dfrac{4r+6}{r(r+1)(r+2)(r+3)}=\dfrac1{r(r+1)}-\dfrac1{(r+2)(r+3)}$ . (3) (b) Hence find, in terms of $n$ , $\displaystyle\sum_{r=1}^{n}\frac{4r+6}{r(r+1)(r+2)(r+3)}$ . (3)
(6)
(Total for Question 2 is 6 marks)
3.
Show that $\displaystyle\sum_{r=1}^{n}\frac1{(r+1)(r+4)}=\frac{n(13n^2+dn+e)}{36(n+2)(n+3)(n+4)}$ , where $d$ and $e$ are integers to be found.
(6)
(Total for Question 3 is 6 marks)
4.
Show that $\dfrac{r}{(r+1)!}=\dfrac1{r!}-\dfrac1{(r+1)!}$ . Hence find $\displaystyle\sum_{r=1}^{n}\dfrac{r}{(r+1)!}$ and determine the least positive integer $n$ for which this sum is greater than $\dfrac{719}{720}$ .
(6)
(Total for Question 4 is 6 marks)

Tier 3 · Hard

1.
Use partial fractions and the method of differences to prove that $\sum_{r=1}^{n}\frac1{r(r+1)(r+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$ . Hence find the corresponding infinite sum.
(6)
(Total for Question 1 is 6 marks)
2.
Show that $\dfrac1{\sqrt{r+1}+\sqrt r}=\sqrt{r+1}-\sqrt r$ ; hence show that $\displaystyle\sum_{r=1}^{n}\dfrac1{\sqrt{r+1}+\sqrt r}=\sqrt{n+1}-1$ , and find the least value of $n$ for which the sum exceeds $20$ .
(6)
(Total for Question 2 is 6 marks)
3.
For each positive integer $r$ , show that $\arctan\left(\dfrac1{r^2+r+1}\right)=\arctan(r+1)-\arctan r$ . Hence find $\displaystyle\sum_{r=1}^{n}\arctan\left(\dfrac1{r^2+r+1}\right)$ and the corresponding sum to infinity.
(7)
(Total for Question 3 is 7 marks)
4.
Let $0<\theta<\dfrac{\pi}{n+1}$ . Show that $\dfrac{\sin\theta}{\sin(r\theta)\sin((r+1)\theta)}=\cot(r\theta)-\cot((r+1)\theta)$ . Hence find $\displaystyle\sum_{r=1}^{n}\dfrac{\sin\theta}{\sin(r\theta)\sin((r+1)\theta)}$ . Evaluate the sum when $n=5$ and $\theta=\pi/12$ .
(7)
(Total for Question 4 is 7 marks)
5.
Show that $\dfrac{2^r}{(2^r+3)(2^{r+1}+3)}=\dfrac1{2^r+3}-\dfrac1{2^{r+1}+3}$ . Hence find a closed form for $\displaystyle\sum_{r=1}^{n}\dfrac{2^r}{(2^r+3)(2^{r+1}+3)}$ , the corresponding sum to infinity, and the exact value of $S_\infty-S_6$ , where $S_n$ is the sum of the first $n$ terms.
(7)
(Total for Question 5 is 7 marks)

CP-4.5 · Find the Maclaurin series of a function including the general term.

Explanation

A Maclaurin series is the Taylor series about zero: $f(x)=\sum_{r=0}^{\infty}\dfrac{f^{(r)}(0)}{r!}x^r$ , within the range of values of $x$ for which the expansion is valid.
Repeated differentiation supplies the derivative pattern and the values at zero.
A complete response should show enough initial terms to establish signs and missing powers, then state a general term with its index range.
Products and compound functions may instead be formed from known series, retaining every contribution up to the required degree; a complex exponential can efficiently generate some trigonometric products.
The factorial belongs to the denominator of every coefficient, and a finite list of terms does not replace a requested general term.

Worked example

Find the Maclaurin series for $\cosh(2x)$ and give its general term.

1.Successive derivatives alternate between multiples of $\cosh(2x)$ and $\sinh(2x)$ .
2.Odd derivatives are zero at zero, while the $2r$ th derivative has value $2^{2r}$ .
3.Substitute these values into the Maclaurin formula.

Answer: $\cosh(2x)=\sum_{r=0}^{\infty}\dfrac{2^{2r}x^{2r}}{(2r)!}=1+2x^2+\dfrac23x^4+\cdots$ .

Common mistakes

Don't fall into the trap of giving several initial terms but no general term when one is explicitly requested.
Don't fall into the trap of using $r$ factorial for a series containing only powers $x^{2r}$ instead of $(2r)!$ .
Don't fall into the trap of including odd powers even though every odd derivative is zero at zero.

Exam tip

List derivative values at zero until the pattern is clear, then state the sigma term and its starting index.

Tier 1 · Easy

1.
Find the Maclaurin series of $f(x)=\frac1{1-x}$ and give its general term.
(3)
(Total for Question 1 is 3 marks)
2.
Find the Maclaurin series of $(1+t)e^t$ up to and including the term in $t^3$ , then state the general term of the series.
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
Find the Maclaurin series of $e^{2t}$ up to and including the term in $t^4$ , and state the general term.
(4)
(Total for Question 1 is 4 marks)
2.
Find the first three non-zero terms and a general term for the Maclaurin series of $\sin(x^2)$ .
(4)
(Total for Question 2 is 4 marks)
3.
By differentiating the geometric series, find a general term for the Maclaurin series of $\dfrac{1+x}{(1-x)^2}$ . Hence find the coefficient of $x^{20}$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Find the Maclaurin series of $e^t\cos t$ up to and including the term in $t^5$ , and give a general term for the series.
(6)
(Total for Question 1 is 6 marks)
2.
For $y=e^{\arcsin t}$ , show that $(1-t^2)y''-ty'-y=0$ and hence find the Maclaurin series for $y$ up to and including the term in $t^4$ .
(6)
(Total for Question 2 is 6 marks)
3.
Find the Maclaurin series of $\ln(1+t+t^2)$ up to and including the term in $t^8$ , and give a general term.
(6)
(Total for Question 3 is 6 marks)
4.
Let $H_n=1+\dfrac12+\cdots+\dfrac1n$ . Find the Maclaurin series of $\dfrac{\ln(1+t)}{1+t}$ up to and including the term in $t^6$ , and give a general term in terms of $H_n$ .
(6)
(Total for Question 4 is 6 marks)
5.
Find the first four non-zero terms and a general term for the Maclaurin series of $\arctan x-\dfrac{x}{1+x^2}$ . State the range of values of $x$ for which the expansion is valid.
(7)
(Total for Question 5 is 7 marks)

CP-4.6 · Recognise and use the Maclaurin series for e^x, ln(1+x), sin x, cos x and (1+x)^n, and be aware of the range of values of x for which they are valid (proof not required).

Explanation

The standard series for $e^x$ , $\sin x$ and $\cos x$ converge for every real $x$ .
The logarithmic series is $\ln(1+x)=x-x^2/2+x^3/3-\cdots$ , valid for $-1<x\le1$ .
The general binomial expansion begins $(1+x)^n=1+nx+n(n-1)x^2/2!+\cdots$ and, when it does not terminate, is valid for $|x|<1$ ; a non-negative integer exponent produces a finite identity valid for all $x$ .
Compound expansions are obtained by substitution, multiplication or division of known series, keeping all terms up to the requested power.
Every substitution also changes the range of values of $x$ for which the expansion is valid, so the original condition must be rewritten.

Worked example

Write the first four non-zero terms of $\ln(1-2x)$ and state the range of values of $x$ for which the expansion is valid.

1.Substitute $u=-2x$ into $\ln(1+u)=u-u^2/2+u^3/3-u^4/4+\cdots$ .
2.This gives $-2x-2x^2-8x^3/3-4x^4+\cdots$ .
3.Transform $-1<u\le1$ into $-1\le2x<1$ .

Answer: $-2x-2x^2-\dfrac83x^3-4x^4+\cdots$ , valid for $-\dfrac12\le x<\dfrac12$ .

Common mistakes

Don't fall into the trap of keeping the original interval after replacing $x$ by a multiple such as $-2x$ .
Don't fall into the trap of writing odd powers in the cosine series or even powers in the sine series.
Don't fall into the trap of treating a non-terminating binomial expansion as valid for all real $x$ .

Exam tip

State the substituted convergence condition alongside the expansion, not as an afterthought.

Tier 1 · Easy

1.
Write down the Maclaurin series for $\cos t$ up to and including the term in $t^6$ , and state the range of values of $t$ for which the expansion is valid.
(3)
(Total for Question 1 is 3 marks)
2.
Use the first four terms of the Maclaurin series for $e^x$ to estimate $e^{0.2}$ . For comparison, the true value is $1.22140$ (5 d.p.). Comment on the accuracy of the estimate.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Write the first four terms of the expansion of $(1-2x)^{-1/2}$ , and state the range of values of $x$ for which it is valid.
(4)
(Total for Question 1 is 4 marks)
2.
Using standard series, expand $e^t(1-t)^{-1/2}$ up to and including the term in $t^3$ . State the range of values of $t$ for which the expansion is valid.
(5)
(Total for Question 2 is 5 marks)
3.
Let $f(u)=(1+4u)^{1/2}-(1-4u)^{1/2}$ . Using standard series, expand $f(u)$ up to and including the term in $u^3$ . State the range of values of $u$ for which the expansion is valid.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Use standard Maclaurin series to show that $\ln\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)$ . State the range of values of $x$ for which the expansion is valid and use terms up to $x^5$ to approximate $\ln(3/2)$ .
(6)
(Total for Question 1 is 6 marks)
2.
Use the Maclaurin series for $\ln(1+t)$ and $\ln(1-t)$ . Determine the expansion of their product up to and including the term in $t^6$ , in ascending powers of $t$ . State the range of values of $t$ for which the expansion is valid. Hence find an estimate for $\ln(4/3)\ln(2/3)$ , giving your answer to $4$ decimal places.
(7)
(Total for Question 2 is 7 marks)
3.
For real $x>-1$ with $x\ne0$ , let $f(x)=\dfrac{\ln(1+x)-\sin x}{x^2}$ . Also, $f(0)$ is defined to be $-\dfrac12$ . Use standard Maclaurin series to expand $f(x)$ up to and including the $x^4$ term. State the range of values of $x$ for which the expansion is valid, and use the expansion to estimate $f(0.1)$ to $4$ decimal places.
(7)
(Total for Question 3 is 7 marks)
4.
Let $k>0$ . Using standard Maclaurin series, expand $\ln(1+kz)\cos z$ up to and including the term in $z^4$ . Given that the coefficient of $z^3$ is zero, determine the value of $k$ and hence give the expansion. State the range of values of $z$ for which the expansion is valid.
(8)
(Total for Question 4 is 8 marks)
5.
The Maclaurin series of $me^{2w}+n\cos w$ has coefficient $5$ for $w^2$ and coefficient $4$ for $w^3$ . Determine the values of $m$ and $n$ . Hence find the expansion up to and including the term in $w^6$ , and state the range of values of $w$ for which the expansion is valid.
(8)
(Total for Question 5 is 8 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

CP-4.1 · Understand and use the relationship between roots and coefficients of polynomial equations up to quartic equations.

Tier 1 · Easy

Mark scheme for CP-4.1 Tier 1 · Easy
Question	Scheme	Marks
1	$5$ , $2$ , $-8$	3
(3 marks)		3
Notes
Compare with $x^3-s_1x^2+s_2x-s_3=0$ . Therefore $s_1=5$ , $s_2=2$ and $-s_3=8$ , so $\alpha\beta\gamma=s_3=-8$ .
2	$s_1=\alpha+\beta+\gamma=4$ , $s_2=\alpha\beta+\beta\gamma+\gamma\alpha=-1$ and $s_3=\alpha\beta\gamma=-6$ $\alpha^2\beta+\alpha\beta^2+\beta^2\gamma+\beta\gamma^2+\gamma^2\alpha+\gamma\alpha^2=s_1s_2-3s_3$ $s_1s_2-3s_3=4(-1)-3(-6)=14$	3
(3 marks)		3
Notes
From the coefficients, $s_1=4$ , $s_2=-1$ and $s_3=-6$ . Expanding $s_1s_2=(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha)$ gives the six required terms together with $3\alpha\beta\gamma$ . Hence the required sum is $s_1s_2-3s_3=4(-1)-3(-6)=14$ .

Tier 2 · Standard

Mark scheme for CP-4.1 Tier 2 · Standard
Question	Scheme	Marks
1	$\frac32$ $\frac74$	4
(4 marks)		4
Notes
After dividing by $2$ , the sum of roots is $-(-3/2)=3/2$ . The sum of triple products is $-7/2$ and the product is $-2$ . Hence the sum of reciprocals is $\frac{\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta}{\alpha\beta\gamma\delta}=\frac{-7/2}{-2}=\frac74$ .
2	$s_1=3$ , $s_2=a$ , $s_3=-b$ and $s_4=6$ $11=s_1^2-2s_2=9-2a$ $a=-1$ $\sum_{i<j}\alpha_i^2\alpha_j^2=s_2^2-2s_1s_3+2s_4$ $25=(-1)^2-2(3)(-b)+2(6)=13+6b$ , so $b=2$	5
(5 marks)		5
Notes
The elementary symmetric sums are $s_1=3$ , $s_2=a$ , $s_3=-b$ and $s_4=6$ . The sum of the root squares gives $11=s_1^2-2s_2=9-2a$ , so $a=-1$ . For four roots, $\sum_{i<j}\alpha_i^2\alpha_j^2=s_2^2-2s_1s_3+2s_4$ . Hence $25=(-1)^2-2(3)(-b)+2(6)=13+6b$ , giving $b=2$ .
3	$f(2)=(2-\alpha)(2-\beta)(2-\gamma)(2-\delta)$ $f(2)=2^4-2(2^3)+3(2^2)+2p+6=18+2p$ $18+2p=10$ , so $p=-4$ $s_2=3$ and $s_4=6$ $\displaystyle\sum_{i<j}\frac1{\alpha_i\alpha_j}=\frac{s_2}{s_4}=\frac12$	5
(5 marks)		5
Notes
For a monic quartic, $f(x)$ is the product of $x$ minus each root. Therefore the given product is $f(2)=18+2p$ , so $p=-4$ . Vieta's formulae give $s_2=3$ and $s_4=6$ . Each reciprocal pair product has the complementary pair product over $s_4$ , so their sum is $s_2/s_4=1/2$ .

Tier 3 · Hard

Mark scheme for CP-4.1 Tier 3 · Hard
Question	Scheme	Marks
1	$p=3$ $\alpha^3+\beta^3+\gamma^3=46$	5
(5 marks)		5
Notes
$\alpha+\beta+\gamma=4$ and $\alpha\beta+\beta\gamma+\gamma\alpha=p$ . Thus $10=4^2-2p$ , giving $p=3$ . Also $\alpha\beta\gamma=6$ . Using $\sum\alpha^3=(\sum\alpha)(\sum\alpha^2)-(\sum\alpha\beta)(\sum\alpha)+3\alpha\beta\gamma$ gives $4(10)-3(4)+3(6)=46$ .
2	$s_1=3$ , $s_2=p$ and $s_3=2$ $s_2^2=\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2+2\alpha\beta\gamma(\alpha+\beta+\gamma)$ $p^2=13+2(2)(3)=25$ $p=5$ since $p>0$ $\dfrac1{\alpha^2}+\dfrac1{\beta^2}+\dfrac1{\gamma^2}=\dfrac{\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2}{(\alpha\beta\gamma)^2}$ $\dfrac1{\alpha^2}+\dfrac1{\beta^2}+\dfrac1{\gamma^2}=\dfrac{13}{4}$	6
(6 marks)		6
Notes
Here $s_1=3$ , $s_2=p$ and $s_3=2$ . Squaring $s_2$ gives $s_2^2=\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2+2\alpha\beta\gamma(\alpha+\beta+\gamma)=13+12=25$ . Since $p>0$ , $p=5$ . Dividing the given squared pair-product sum by $(\alpha\beta\gamma)^2=4$ gives the reciprocal-square sum $13/4$ .
3	$s_1=4$ , $s_2=m$ , $s_3=-n$ and $s_4=3$ $\displaystyle\sum_{i<j}(r_i-r_j)^2=3s_1^2-8s_2=48-8m$ $48-8m=64$ , so $m=-2$ The second given expression is $s_1s_3-4s_4$ $-40=4(-n)-4(3)$ , so $n=7$ The required sum is $s_3^2-2s_2s_4$ $s_3^2-2s_2s_4=(-7)^2-2(-2)(3)=61$	7
(7 marks)		7
Notes
Here $s_1=4$ , $s_2=m$ , $s_3=-n$ and $s_4=3$ . For four roots, $\sum_{i<j}(r_i-r_j)^2=3s_1^2-8s_2=48-8m$ . Hence $m=-2$ . Expanding $s_1s_3$ shows that the second given expression is $s_1s_3-4s_4$ , so $-40=-4n-12$ and $n=7$ . The four triple products have sum $s_3$ and pair-product sum $s_2s_4$ , so the required sum of their squares is $s_3^2-2s_2s_4=61$ .
4	$s_1=6$ and $s_2=p$ $74=s_1^2-2s_2=36-2p$ $p=-19$ Substitution of the root $-5$ gives $625+750-475-5q-180=0$ $q=144$ $x^4-6x^3-19x^2+144x-180=(x+5)(x^3-11x^2+36x-36)$ $(x+5)(x-2)(x-3)(x-6)$ , so the other roots are $2$ , $3$ and $6$	7
(7 marks)		7
Notes
Vieta's formulae give $s_1=6$ and $s_2=p$ . Hence the square-sum condition is $74=6^2-2p$ , giving $p=-19$ . Substitution of the known root $x=-5$ then gives $q=144$ . Dividing by $x+5$ gives $x^3-11x^2+36x-36=(x-2)(x-3)(x-6)$ , so the remaining roots are $2$ , $3$ and $6$ .
5	$s_1=5$ , $s_2=7$ and $s_3=3$ $\alpha\beta+\beta\gamma+\gamma\alpha=7$ $\alpha\beta^2\gamma+\beta\gamma^2\alpha+\gamma\alpha^2\beta=s_1s_3=15$ $(\alpha\beta)(\beta\gamma)(\gamma\alpha)=s_3^2=9$ The required equation is $u^3-7u^2+15u-9=0$ The numerator of the shifted reciprocal sum is $15-4(7)+12=-1$ Its denominator is $9-2(15)+4(7)-8=-1$ , so the sum is $1$	7
(7 marks)		7
Notes
The pair-product roots have sum $s_2=7$ , pair-product sum $s_1s_3=15$ and product $s_3^2=9$ , giving $u^3-7u^2+15u-9=0$ . If these roots are $U,V,W$ , then $\sum1/(U-2)$ has numerator $UV+UW+VW-4(U+V+W)+12=-1$ and denominator $UVW-2(UV+UW+VW)+4(U+V+W)-8=-1$ . The required value is therefore $1$ .

CP-4.2 · Form a polynomial equation whose roots are a linear transformation of the roots of a given polynomial equation (of at least cubic degree).

Tier 1 · Easy

Mark scheme for CP-4.2 Tier 1 · Easy
Question	Scheme	Marks
1	$x^3-6x^2+9x-1=0$	3
(3 marks)		3
Notes
Let a new root be $x=t+2$ , so $t=x-2$ . Substitute into the original equation: $(x-2)^3-3(x-2)+1=0$ . Expanding and collecting terms gives $x^3-6x^2+9x-1=0$ .
2	$w=2x-1$ $(2x-1)^3+3(2x-1)^2-4(2x-1)-12=0$ $8x^3-14x-6=0$ $4x^3-7x-3=0$	4
(4 marks)		4
Notes
For an original root $x$ , the corresponding given root is $w=2x-1$ . Substitution gives $(2x-1)^3+3(2x-1)^2-4(2x-1)-12=0$ . Expanding yields $8x^3-14x-6=0$ , which simplifies to $4x^3-7x-3=0$ .

Tier 2 · Standard

Mark scheme for CP-4.2 Tier 2 · Standard
Question	Scheme	Marks
1	$2x^3+3x^2+36x-46=0$	4
(4 marks)		4
Notes
Let $x=3t-1$ , so $t=(x+1)/3$ . Substitute into the original polynomial and multiply by $27$ : $2(x+1)^3-3(x+1)^2+36(x+1)-81=0$ . Expanding gives $2x^3+3x^2+36x-46=0$ .
2	$t=(u-3)/2$ $(u-3)^3-4(u-3)^2+20(u-3)-8=0$ $u^3-13u^2+71u-131=0$ , so $d=-13$ , $e=71$ and $f=-131$ $u=1-x$ $(1-x)^3-13(1-x)^2+71(1-x)-131=-x^3-10x^2-48x-72=0$ ; multiplying throughout by $-1$ gives $x^3+10x^2+48x+72=0$	5
(5 marks)		5
Notes
For part (a), set $u=2t+3$ , so $t=(u-3)/2$ . Substitution followed by multiplication by $8$ gives $(u-3)^3-4(u-3)^2+20(u-3)-8=0$ , or $u^3-13u^2+71u-131=0$ . Thus $d=-13$ , $e=71$ and $f=-131$ . For part (b), set $x=1-u$ , so $u=1-x$ . Substitution into the equation from part (a) gives $-x^3-10x^2-48x-72=0$ ; multiplying throughout by $-1$ gives $x^3+10x^2+48x+72=0$ .
3	$\alpha+\beta+\gamma=5$ $\alpha+2\beta+2\gamma=10-\alpha$ , with the corresponding cyclic results For a new root $x$ , set $x=10-t$ , so $t=10-x$ $(10-x)^3-5(10-x)^2+2(10-x)+7=0$ $x^3-25x^2+202x-527=0$	5
(5 marks)		5
Notes
The original roots sum to $5$ . Therefore $\alpha+2\beta+2\gamma=2(\alpha+\beta+\gamma)-\alpha=10-\alpha$ , and similarly the other two required roots are $10-\beta$ and $10-\gamma$ . Set $x=10-t$ , so $t=10-x$ . Substitution into the given equation gives $(10-x)^3-5(10-x)^2+2(10-x)+7=0$ . Expansion and multiplication by $-1$ give $x^3-25x^2+202x-527=0$ .

Tier 3 · Hard

Mark scheme for CP-4.2 Tier 3 · Hard
Question	Scheme	Marks
1	$x^4-2x^3-3x^2-77x+85=0$	6
(6 marks)		6
Notes
Let $x=2-3t$ , so $t=(2-x)/3$ . Substitute into the original polynomial and multiply by $3^4=81$ : $(2-x)^4-6(2-x)^3+9(2-x)^2+81(2-x)-81=0$ . Expanding gives $x^4-2x^3-3x^2-77x+85=0$ .
2	$\alpha+\beta+\gamma=3$ and $\alpha\beta+\beta\gamma+\gamma\alpha=2$ $3m+3n=-3$ , so $m+n=-1$ $2m^2+6mn+3n^2=-1$ $3-m^2=-1$ $m=2$ and $n=-3$ $t=(x+3)/2$ $d=-43$	7
(7 marks)		7
Notes
The original symmetric sums are $s_1=3$ and $s_2=2$ . The new sum is $3m+3n=-3$ , so $m+n=-1$ . The new pair-product sum is $2m^2+2mn(3)+3n^2=-1$ . Using $n=-1-m$ reduces this to $3-m^2=-1$ , hence $m=2$ because $m>0$ , and $n=-3$ . Thus $x=2t-3$ , so $t=(x+3)/2$ . Substitution and multiplication by $8$ give $(x+3)^3-6(x+3)^2+8(x+3)-40=0$ , which expands to $x^3+3x^2-x-43=0$ , so $d=-43$ .
3	$\alpha+\beta+\gamma+\delta=4$ $4+4h=0$ , so $h=-1$ Set $u=t-1$ , so $t=u+1$ $2(u+1)^4-8(u+1)^3+3(u+1)^2+5(u+1)-7=0$ $2u^4-9u^2-5u-5=0$ Set $x=2u-5$ , so $u=(x+5)/2$ $2(x+5)^4-36(x+5)^2-40(x+5)-80=0$ $x^4+20x^3+132x^2+300x+35=0$	8
(8 marks)		8
Notes
The original roots sum to $4$ . The shifted roots therefore sum to $4+4h$ , giving $h=-1$ . If a shifted root is $u=t-1$ , then $t=u+1$ . Substitution and expansion give the depressed quartic $2u^4-9u^2-5u-5=0$ . For part (b), write $x=2u-5$ , so $u=(x+5)/2$ . Substitution and multiplication by $16$ give $2(x+5)^4-36(x+5)^2-40(x+5)-80=0$ . Expansion and division by $2$ give $x^4+20x^3+132x^2+300x+35=0$ .
4	From $t=3\alpha+1$ , $\alpha=(t-1)/3$ For a new root $u$ , $u=2\alpha-5=2(t-1)/3-5=(2t-17)/3$ $t=(3u+17)/2$ $2[(3u+17)/2]^4-3[(3u+17)/2]^3+5[(3u+17)/2]^2-4[(3u+17)/2]+7=0$ Multiplying by $8$ gives $(3u+17)^4-3(3u+17)^3+10(3u+17)^2-16(3u+17)+56=0$ $81u^4+1755u^3+14319u^2+52125u+71456=0$ $\gcd(81,1755,14319,52125,71456)=1$ , and the leading coefficient is positive The sum of the four new roots is $-1755/81=-65/3$	8
(8 marks)		8
Notes
If $t=3\alpha+1$ , then $\alpha=(t-1)/3$ . Therefore the corresponding new root is $u=2\alpha-5=(2t-17)/3$ , so $t=(3u+17)/2$ . Substitution into the given quartic and multiplication by $8$ give $(3u+17)^4-3(3u+17)^3+10(3u+17)^2-16(3u+17)+56=0$ . This expands to the primitive equation $81u^4+1755u^3+14319u^2+52125u+71456=0$ . By Vieta's formula, the sum of its roots is $-1755/81=-65/3$ .
5	$\alpha+\beta+\gamma=4$ $2\alpha-\beta-\gamma=3\alpha-4$ , with the corresponding results for the other roots For a new root $u$ , $u=3t-4$ , so $t=(u+4)/3$ $27[(u+4)/3]^3-108[(u+4)/3]^2+27[(u+4)/3]+54=0$ The equation simplifies to $u^3-39u-38=0$ For the new roots, $S_1=0$ , $S_2=-39$ and $S_3=38$ Their square sum is $S_1^2-2S_2=78$ Their shifted product is $S_3+2S_2+4S_1+8=38-78+8=-32$	8
(8 marks)		8
Notes
Since the original roots sum to $4$ , each stated root is $3t-4$ . Put $t=(u+4)/3$ in the original equation and multiply by $27$ to obtain $u^3-39u-38=0$ . The new symmetric sums are $S_1=0$ , $S_2=-39$ and $S_3=38$ . Hence the square sum is $78$ , while $(u_1+2)(u_2+2)(u_3+2)=S_3+2S_2+4S_1+8=-32$ .

CP-4.3 · Understand and use formulae for the sums of integers, squares and cubes and use these to sum other series.

Tier 1 · Easy

Mark scheme for CP-4.3 Tier 1 · Easy
Question	Scheme	Marks
1	$3080$	3
(3 marks)		3
Notes
$r(r+1)=r^2+r$ , so the sum is $\frac{20(21)(41)}6+\frac{20(21)}2=2870+210=3080$ .
2	$\displaystyle\sum_{r=3}^{10}r^2=\dfrac{10(11)(21)}6-(1^2+2^2)=380$ $\displaystyle\sum_{r=3}^{10}r=\dfrac{10(11)}2-(1+2)=52$ $\displaystyle\sum_{r=3}^{10}(r^2-r)=380-52=328$	3
(3 marks)		3
Notes
Use initial sums and remove the first two terms. The square sum is $10(11)(21)/6-(1^2+2^2)=380$ , while the integer sum is $10(11)/2-(1+2)=52$ . Their difference is $380-52=328$ .

Tier 2 · Standard

Mark scheme for CP-4.3 Tier 2 · Standard
Question	Scheme	Marks
1	$\frac{n(2n^2+n+7)}2$	4
(4 marks)		4
Notes
Use standard sums: $3\sum r^2-2\sum r+4n=\frac{n(n+1)(2n+1)}2-n(n+1)+4n$ . Factoring $n/2$ gives $\frac n2[(n+1)(2n+1)-2(n+1)+8]=\frac{n(2n^2+n+7)}2$ .
2	$\displaystyle\sum_{r=1}^{8}r^2=\dfrac{8(9)(17)}6=204$ $\displaystyle\sum_{r=1}^{8}r=\dfrac{8(9)}2=36$ $204+36k=420$ $k=6$	4
(4 marks)		4
Notes
$\sum_{r=1}^{8}r^2=8(9)(17)/6=204$ and $\sum_{r=1}^{8}r=8(9)/2=36$ . Therefore $204+36k=420$ , so $36k=216$ and $k=6$ .
3	$\displaystyle\sum_{r=1}^{n}(3r^2-r)=3\sum r^2-\sum r$ $\displaystyle=\frac{n(n+1)(2n+1)}2-\frac{n(n+1)}2$ $\displaystyle\sum_{r=1}^{n}(3r^2-r)=n^2(n+1)$ $n^2(n+1)=576=8^2(8+1)$ Since $n^2(n+1)$ is strictly increasing for positive integers, $n=8$	5
(5 marks)		5
Notes
Split the sum as $3\sum r^2-\sum r$ . Substitution of the standard results gives $n(n+1)(2n+1)/2-n(n+1)/2=n^2(n+1)$ . The equation is therefore $n^2(n+1)=576=8^2(9)$ . This expression increases strictly for positive integers, so the unique solution is $n=8$ .

Tier 3 · Hard

Mark scheme for CP-4.3 Tier 3 · Hard
Question	Scheme	Marks
1	$\frac{n(n+1)^2(n+2)}2$ $32640$	6
(6 marks)		6
Notes
Expand $r(r+1)(2r+1)=2r^3+3r^2+r$ . Therefore the sum is $2[\frac{n(n+1)}2]^2+3\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2$ . Factoring and simplifying gives $\frac{n(n+1)^2(n+2)}2$ . At $n=15$ , this is $\frac{15(16)^2(17)}2=32640$ .
2	$5\sum r^2-3\sum r+2k$ $\dfrac{5k(k+1)(2k+1)}6-\dfrac{3k(k+1)}2+2k$ $\dfrac{k}{6}[5(k+1)(2k+1)-9(k+1)+12]$ $\displaystyle\sum_{r=1}^{k}(5r^2-3r+2)=\dfrac{k(5k^2+3k+4)}3$ $S_n=\dfrac13\left(5\sum k^3+3\sum k^2+4\sum k\right)$ $S_n=\dfrac13\left(5\left[\dfrac{n(n+1)}2\right]^2+\dfrac{3n(n+1)(2n+1)}6+\dfrac{4n(n+1)}2\right)$ $S_n=\dfrac{n(n+1)}{12}[5n(n+1)+2(2n+1)+8]$ $S_n=\dfrac{n(n+1)(5n^2+9n+10)}{12}$ $S_{25}=182000$ and $S_{10}=5500$ $S_{25}-S_{10}=176500$	10
(10 marks)		10
Notes
For part (a), use $5\sum r^2-3\sum r+2k$ and simplify to $k(5k^2+3k+4)/3$ . Therefore $S_n=\frac13(5\sum k^3+3\sum k^2+4\sum k)$ . Substitution of the standard formulae and factorisation give $S_n=n(n+1)(5n^2+9n+10)/12$ . The closed form gives $S_{25}=182000$ and $S_{10}=5500$ , so the required difference is $176500$ .
3	$11a+3b+c=49$ from $S_5=245$ $51a+9b+2c=231$ from $S_8=924$ $64a+8b+c=294$ $53a+5b=245$ $29a+3b=133$ $a=5$ $b=-4$ and $c=6$ $S_n=5\dfrac{n(n+1)(2n+1)}6-4\dfrac{n(n+1)}2+6n=\dfrac{n(10n^2+3n+29)}6$ $S_{18}=9969$ and $S_6=407$ $\displaystyle\sum_{r=7}^{18}(5r^2-4r+6)=S_{18}-S_6=9562$	10
(10 marks)		10
Notes
Using the sums to $5$ and $8$ , and simplifying, gives $11a+3b+c=49$ and $51a+9b+2c=231$ . The eighth term gives $64a+8b+c=294$ . Eliminating $c$ gives $53a+5b=245$ and $29a+3b=133$ , so $a=5$ , $b=-4$ and $c=6$ . The standard sums then give $S_n=5n(n+1)(2n+1)/6-4n(n+1)/2+6n=n(10n^2+3n+29)/6$ . Hence the required block is $S_{18}-S_6=9969-407=9562$ .
4	$\displaystyle\sum_{r=1}^{n-1}r(n-r)=n\sum_{r=1}^{n-1}r-\sum_{r=1}^{n-1}r^2$ $\displaystyle=\frac{n^2(n-1)}2-\frac{n(n-1)(2n-1)}6$ $\displaystyle=\frac{n(n-1)}6[3n-(2n-1)]$ $\displaystyle\sum_{r=1}^{n-1}r(n-r)=\frac{n(n-1)(n+1)}6$ $\displaystyle T_N=\frac16\sum_{n=2}^{N}(n^3-n)$ $\displaystyle T_N=\frac16\left(\left[\frac{N(N+1)}2\right]^2-\frac{N(N+1)}2\right)$ $\displaystyle T_N=\frac{N(N+1)}{24}[N(N+1)-2]$ $\displaystyle T_N=\frac{N(N-1)(N+1)(N+2)}{24}$ $\displaystyle T_{12}=\frac{12(11)(13)(14)}{24}$ $T_{12}=1001$	10
(10 marks)		10
Notes
For the inner sum, use $n\sum r-\sum r^2$ with upper limit $n-1$ and simplify to $n(n-1)(n+1)/6$ . Summing this from $n=2$ to $N$ gives $\frac16(\sum n^3-\sum n)$ . The $n=1$ contribution is zero, so the standard formulae give $N(N+1)[N(N+1)-2]/24=N(N-1)(N+1)(N+2)/24$ . At $N=12$ this is $1001$ .
5	$\displaystyle\sum_{r=1}^{n}(3r+2)^2=9\sum r^2+12\sum r+4n$ $\displaystyle=\frac{3n(n+1)(2n+1)}2+6n(n+1)+4n$ $\displaystyle=\frac n2[3(n+1)(2n+1)+12(n+1)+8]$ $\displaystyle\sum_{r=1}^{n}(3r+2)^2=\frac{n(6n^2+21n+23)}2$ The mean of the arithmetic sequence is $[5+(3n+2)]/2=(3n+7)/2$ Writing $x_r=3r+2$ and $\bar{x}=(3n+7)/2$ , the squared-deviation sum is $\sum x_r^2-n\bar{x}^2$ $\displaystyle=\frac{n(6n^2+21n+23)}2-\frac{n(3n+7)^2}{4}$ $\displaystyle=\frac n4[12n^2+42n+46-(9n^2+42n+49)]$ The sum of squared deviations is $\dfrac{3n(n^2-1)}4$ For $n=12$ , the sum is $3(12)(12^2-1)/4=1287$	10
(10 marks)		10
Notes
Expand $(3r+2)^2$ and use the standard sums to obtain $n(6n^2+21n+23)/2$ ; at $n=1$ both sides equal $25$ . The arithmetic sequence has mean $(3n+7)/2$ . Using $\sum(x_r-\bar{x})^2=\sum x_r^2-n\bar{x}^2$ then gives $3n(n^2-1)/4$ , which is $1287$ when $n=12$ .

CP-4.4 · Understand and use the method of differences for summation of series including use of partial fractions.

Tier 1 · Easy

Mark scheme for CP-4.4 Tier 1 · Easy
Question	Scheme	Marks
1	$\frac n{3(n+3)}$	3
(3 marks)		3
Notes
$\frac1{(r+2)(r+3)}=\frac1{r+2}-\frac1{r+3}$ . Hence the sum is $(\frac13-\frac14)+(\frac14-\frac15)+\cdots+(\frac1{n+2}-\frac1{n+3})=\frac13-\frac1{n+3}=\frac n{3(n+3)}$ .

Tier 2 · Standard

Mark scheme for CP-4.4 Tier 2 · Standard
Question	Scheme	Marks
1	$\frac{n(3n+5)}{4(n+1)(n+2)}$ $\frac{175}{264}$	6
(6 marks)		6
Notes
$\frac1{r(r+2)}=\frac12(\frac1r-\frac1{r+2})$ . After cancellation, the sum is $\frac12(1+\frac12-\frac1{n+1}-\frac1{n+2})=\frac{n(3n+5)}{4(n+1)(n+2)}$ . Substituting $n=10$ gives $\frac{10(35)}{4(11)(12)}=\frac{175}{264}$ .
2	$\dfrac1{r(r+1)}-\dfrac1{(r+2)(r+3)}=\dfrac{(r+2)(r+3)-r(r+1)}{r(r+1)(r+2)(r+3)}$ $(r+2)(r+3)-r(r+1)=4r+6$ $\dfrac{4r+6}{r(r+1)(r+2)(r+3)}=\dfrac1{r(r+1)}-\dfrac1{(r+2)(r+3)}$ With $A_r=1/[r(r+1)]$ , $\displaystyle\sum_{r=1}^{n}(A_r-A_{r+2})=A_1+A_2-A_{n+1}-A_{n+2}$ $\dfrac12+\dfrac16-\dfrac1{(n+1)(n+2)}-\dfrac1{(n+2)(n+3)}$ $\dfrac{2n(n+4)}{3(n+1)(n+3)}$	6
(6 marks)		6
Notes
The summand is $1/[r(r+1)]-1/[(r+2)(r+3)]$ . With $A_r=1/[r(r+1)]$ , the sum is $A_1+A_2-A_{n+1}-A_{n+2}$ . Thus it equals $1/2+1/6-1/[(n+1)(n+2)]-1/[(n+2)(n+3)]$ , which simplifies to $2n(n+4)/[3(n+1)(n+3)]$ .
3	$\dfrac1{(r+1)(r+4)}=\dfrac13\left(\dfrac1{r+1}-\dfrac1{r+4}\right)$ $\dfrac13\left(\dfrac12+\dfrac13+\dfrac14-\dfrac1{n+2}-\dfrac1{n+3}-\dfrac1{n+4}\right)$ $\dfrac13\left[\dfrac{13}{12}-\dfrac{3n^2+18n+26}{(n+2)(n+3)(n+4)}\right]$ $\dfrac{n(13n^2+81n+122)}{36(n+2)(n+3)(n+4)}$ $d=81$ $e=122$	6
(6 marks)		6
Notes
$1/[(r+1)(r+4)]=\frac13[1/(r+1)-1/(r+4)]$ . Writing the boundary terms after cancellation gives $\frac13[1/2+1/3+1/4-1/(n+2)-1/(n+3)-1/(n+4)]$ . Combining these fractions gives $n(13n^2+81n+122)/[36(n+2)(n+3)(n+4)]$ .
4	$\dfrac1{r!}-\dfrac1{(r+1)!}=\dfrac{r+1-1}{(r+1)!}$ $\dfrac1{r!}-\dfrac1{(r+1)!}=\dfrac r{(r+1)!}$ $\displaystyle\sum_{r=1}^{n}\left(\frac1{r!}-\frac1{(r+1)!}\right)$ $\displaystyle=1-\frac1{(n+1)!}$ $1-\dfrac1{(n+1)!}>\dfrac{719}{720}$ requires $(n+1)!>720$ $n=5$ gives equality because $(n+1)!=720$ , whereas $n=6$ gives $(n+1)!=5040>720$ , so the least possible value is $n=6$	6
(6 marks)		6
Notes
Using $(r+1)!=(r+1)r!$ gives $1/r!-1/(r+1)!=r/(r+1)!$ . The resulting series cancels to $1-1/(n+1)!$ . For this to exceed $719/720$ , $(n+1)!$ must exceed $720$ . At $n=5$ it equals $720$ , while at $n=6$ it is $5040$ , so the least value is $6$ .

Tier 3 · Hard

Mark scheme for CP-4.4 Tier 3 · Hard
Question	Scheme	Marks
1	$\frac{n(n+3)}{4(n+1)(n+2)}$ Infinite sum $=\frac14$	6
(6 marks)		6
Notes
$\frac1{r(r+1)(r+2)}=\frac12[\frac1{r(r+1)}-\frac1{(r+1)(r+2)}]$ . The finite sum is therefore $\frac12[\frac1{1\cdot2}-\frac1{(n+1)(n+2)}]=\frac14-\frac1{2(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$ . Letting $n\to\infty$ gives $\frac14$ .
2	$\dfrac1{\sqrt{r+1}+\sqrt r}\times\dfrac{\sqrt{r+1}-\sqrt r}{\sqrt{r+1}-\sqrt r}=\dfrac{\sqrt{r+1}-\sqrt r}{(r+1)-r}$ $\dfrac1{\sqrt{r+1}+\sqrt r}=\sqrt{r+1}-\sqrt r$ $\displaystyle\sum_{r=1}^{n}(\sqrt{r+1}-\sqrt r)=(\sqrt2-1)+(\sqrt3-\sqrt2)+\cdots+(\sqrt{n+1}-\sqrt n)$ $\displaystyle\sum_{r=1}^{n}\dfrac1{\sqrt{r+1}+\sqrt r}=\sqrt{n+1}-1$ $\sqrt{n+1}-1>20$ gives $\sqrt{n+1}>21$ , so $n>440$ $n=440$ gives a sum of $20$ , so the least value for which the sum exceeds $20$ is $n=441$	6
(6 marks)		6
Notes
Multiplying numerator and denominator by $\sqrt{r+1}-\sqrt r$ gives the stated difference. The finite sum is therefore $(\sqrt2-1)+(\sqrt3-\sqrt2)+\cdots+(\sqrt{n+1}-\sqrt n)=\sqrt{n+1}-1$ . For this to exceed $20$ , $\sqrt{n+1}>21$ , so $n>440$ . At $n=440$ the sum equals $20$ , while at $n=441$ it is $\sqrt{442}-1>20$ ; hence the least value is $441$ .
3	$\displaystyle\tan(\arctan(r+1)-\arctan r)=\frac{(r+1)-r}{1+r(r+1)}$ $\displaystyle\tan(\arctan(r+1)-\arctan r)=\frac1{r^2+r+1}$ Both sides of the proposed identity are angles in $(0,\pi/2)$ , so the equality follows without an added multiple of $\pi$ $\displaystyle\sum_{r=1}^{n}[\arctan(r+1)-\arctan r]$ cancels in its intermediate terms $\displaystyle\sum_{r=1}^{n}\arctan\left(\frac1{r^2+r+1}\right)=\arctan(n+1)-\frac\pi4$ $\displaystyle\lim_{n\to\infty}\arctan(n+1)=\frac\pi2$ The sum to infinity is $\dfrac\pi4$	7
(7 marks)		7
Notes
The tangent subtraction formula gives $\tan(\arctan(r+1)-\arctan r)=1/[1+r(r+1)]$ . Both the difference and $\arctan(1/(r^2+r+1))$ lie in $(0,\pi/2)$ , so they are equal. Summing the differences cancels every intermediate arctangent and leaves $\arctan(n+1)-\arctan1=\arctan(n+1)-\pi/4$ . As $n$ tends to infinity this approaches $\pi/2-\pi/4=\pi/4$ .
4	$\displaystyle\cot(r\theta)-\cot((r+1)\theta)=\frac{\cos(r\theta)\sin((r+1)\theta)-\sin(r\theta)\cos((r+1)\theta)}{\sin(r\theta)\sin((r+1)\theta)}$ The numerator is $\sin((r+1)\theta-r\theta)=\sin\theta$ $\displaystyle\frac{\sin\theta}{\sin(r\theta)\sin((r+1)\theta)}=\cot(r\theta)-\cot((r+1)\theta)$ $\displaystyle\sum_{r=1}^{n}[\cot(r\theta)-\cot((r+1)\theta)]$ The intermediate cotangent terms cancel $\displaystyle\sum_{r=1}^{n}\frac{\sin\theta}{\sin(r\theta)\sin((r+1)\theta)}=\cot\theta-\cot((n+1)\theta)$ For $n=5$ and $\theta=\pi/12$ , the value is $\cot(\pi/12)-\cot(\pi/2)=2+\sqrt3$	7
(7 marks)		7
Notes
Writing the cotangent difference over a common denominator gives numerator $\sin((r+1)\theta-r\theta)=\sin\theta$ . Summing the resulting differences cancels every intermediate cotangent and leaves $\cot\theta-\cot((n+1)\theta)$ . At the stated values this is $\cot(\pi/12)-\cot(\pi/2)=2+\sqrt3$ .
5	$\displaystyle\frac1{2^r+3}-\frac1{2^{r+1}+3}=\frac{2^{r+1}+3-(2^r+3)}{(2^r+3)(2^{r+1}+3)}$ The numerator is $2^r$ , proving the identity $\displaystyle\sum_{r=1}^{n}\left(\frac1{2^r+3}-\frac1{2^{r+1}+3}\right)$ $\displaystyle=\frac15-\frac1{2^{n+1}+3}$ after cancellation $\displaystyle S_n=\frac15-\frac1{2^{n+1}+3}$ $\displaystyle S_\infty=\frac15$ $\displaystyle S_\infty-S_6=\frac1{2^7+3}=\frac1{131}$	7
(7 marks)		7
Notes
The difference of the two reciprocal terms has numerator $2^{r+1}-2^r=2^r$ , so the identity follows. The finite series cancels to $1/5-1/(2^{n+1}+3)$ . Its limit is $1/5$ , and subtracting the six-term partial sum leaves $1/(2^7+3)=1/131$ .

CP-4.5 · Find the Maclaurin series of a function including the general term.

Tier 1 · Easy

Mark scheme for CP-4.5 Tier 1 · Easy
Question	Scheme	Marks
1	$1+x+x^2+x^3+\cdots=\sum_{r=0}^{\infty}x^r$	3
(3 marks)		3
Notes
$f^{(r)}(x)=r!(1-x)^{-(r+1)}$ , so $f^{(r)}(0)=r!$ . Hence the coefficient of $x^r$ is $f^{(r)}(0)/r!=1$ , giving $\sum_{r=0}^{\infty}x^r$ .
2	$e^t=1+t+\dfrac{t^2}{2!}+\dfrac{t^3}{3!}+\cdots$ Multiplying by $1+t$ gives $1+2t+\dfrac32t^2+\dfrac23t^3+\cdots$ For $r\ge1$ , the coefficient of $t^r$ is $\dfrac1{r!}+\dfrac1{(r-1)!}=\dfrac{r+1}{r!}$ $\displaystyle\sum_{r=0}^{\infty}\dfrac{r+1}{r!}t^r$	4
(4 marks)		4
Notes
Multiply $(1+t)$ by $\sum_{r=0}^{\infty}t^r/r!$ . For $r\ge1$ , the coefficient of $t^r$ is $1/r!+1/(r-1)!=(r+1)/r!$ . This gives $1+2t+3t^2/2+2t^3/3+\cdots$ and the stated general term.

Tier 2 · Standard

Mark scheme for CP-4.5 Tier 2 · Standard
Question	Scheme	Marks
1	$1+2t+2t^2+\frac43t^3+\frac23t^4+\cdots$ General term $\frac{2^r t^r}{r!}$	4
(4 marks)		4
Notes
For $f(t)=e^{2t}$ , $f^{(r)}(t)=2^r e^{2t}$ , so $f^{(r)}(0)=2^r$ . Thus $e^{2t}=\sum_{r=0}^{\infty}\frac{2^r t^r}{r!}=1+2t+\frac{4t^2}{2!}+\frac{8t^3}{3!}+\frac{16t^4}{4!}+\cdots$ , which simplifies to the stated expansion.
2	$\sin u=u-\dfrac{u^3}{3!}+\dfrac{u^5}{5!}-\cdots$ Substitute $u=x^2$ $x^2-\dfrac{x^6}{3!}+\dfrac{x^{10}}{5!}+\cdots$ $\displaystyle\sum_{r=0}^{\infty}\frac{(-1)^r x^{4r+2}}{(2r+1)!}$	4
(4 marks)		4
Notes
Substitute $x^2$ for the argument in $\sin u=\sum_{r=0}^{\infty}(-1)^ru^{2r+1}/(2r+1)!$ . Since $(x^2)^{2r+1}=x^{4r+2}$ , the series is $\sum_{r=0}^{\infty}(-1)^rx^{4r+2}/(2r+1)!$ , whose first terms are $x^2-x^6/3!+x^{10}/5!$ .
3	$\displaystyle\frac1{1-x}=\sum_{r=0}^{\infty}x^r$ $\displaystyle\frac1{(1-x)^2}=\sum_{r=0}^{\infty}(r+1)x^r$ $\displaystyle\frac{1+x}{(1-x)^2}=\sum_{r=0}^{\infty}(2r+1)x^r$ for $\|x\|<1$ The coefficient of $x^{20}$ is $2(20)+1=41$	4
(4 marks)		4
Notes
Differentiate the geometric series to get $(1-x)^{-2}=\sum_{r=0}^{\infty}(r+1)x^r$ . Multiplication by $1+x$ gives coefficient $1$ for $x^0$ and $(r+1)+r=2r+1$ for $x^r$ when $r\ge1$ . Thus the general term is $(2r+1)x^r$ , and the coefficient at $r=20$ is $41$ .

Tier 3 · Hard

Mark scheme for CP-4.5 Tier 3 · Hard
Question	Scheme	Marks
1	$1+t-\frac{t^3}{3}-\frac{t^4}{6}-\frac{t^5}{30}+\cdots$ General term $\frac{2^{r/2}\cos(r\pi/4)}{r!}t^r$	6
(6 marks)		6
Notes
$e^t\cos t$ is the real part of $e^{(1+i)t}$ . Since $1+i=\sqrt2e^{i\pi/4}$ , the real part of the $r$ th term $\frac{(1+i)^r t^r}{r!}$ is $\frac{2^{r/2}\cos(r\pi/4)}{r!}t^r$ . Substituting $r=0,1,2,3,4,5$ gives $1+t+0t^2-\frac{t^3}{3}-\frac{t^4}{6}-\frac{t^5}{30}+\cdots$ .
2	$y'=y/\sqrt{1-t^2}$ $y''=y/(1-t^2)+ty/(1-t^2)^{3/2}$ $(1-t^2)y''-ty'-y=0$ $y(0)=1$ and $y'(0)=1$ $a_2=1/2$ , $a_3=1/3$ and $a_4=5/24$ $y=1+t+\dfrac{t^2}{2}+\dfrac{t^3}{3}+\dfrac{5t^4}{24}+\cdots$	6
(6 marks)		6
Notes
Differentiation gives $y'=y/\sqrt{1-t^2}$ and $y''=y/(1-t^2)+ty/(1-t^2)^{3/2}$ , which proves the differential equation. Put $y=\sum_{r=0}^{\infty}a_rt^r$ . Since $y(0)=1$ and $y'(0)=1$ , $a_0=a_1=1$ . Equating coefficients in the differential equation gives $a_2=1/2$ , $a_3=1/3$ and $a_4=5/24$ , producing the stated series.
3	$1+t+t^2=(1-t^3)/(1-t)$ $\ln(1+t+t^2)=\ln(1-t^3)-\ln(1-t)$ $\displaystyle\ln(1-t^3)=-\sum_{r=1}^{\infty}\frac{t^{3r}}r$ $\displaystyle-\ln(1-t)=\sum_{r=1}^{\infty}\frac{t^r}r$ The coefficients of $t^3$ and $t^6$ are $-2/3$ and $-1/3$ respectively, giving $t+\dfrac{t^2}{2}-\dfrac{2t^3}{3}+\dfrac{t^4}{4}+\dfrac{t^5}{5}-\dfrac{t^6}{3}+\dfrac{t^7}{7}+\dfrac{t^8}{8}+\cdots$ $\displaystyle\ln(1+t+t^2)=\sum_{m=1}^{\infty}c_mt^m$ , where $c_m=-2/m$ when $m$ is a multiple of $3$ and $c_m=1/m$ when $m$ is not a multiple of $3$ , for $\|t\|<1$	6
(6 marks)		6
Notes
Factor $1+t+t^2=(1-t^3)/(1-t)$ . Hence its logarithm is $\ln(1-t^3)-\ln(1-t)=-\sum_{r=1}^{\infty}t^{3r}/r+\sum_{r=1}^{\infty}t^r/r$ . Combining all contributions up to $t^8$ gives $t+t^2/2-2t^3/3+t^4/4+t^5/5-t^6/3+t^7/7+t^8/8+\cdots$ . Thus the coefficient is $1/m$ unless $m$ is a multiple of $3$ , when the second series changes it to $-2/m$ .
4	$\displaystyle\ln(1+t)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}k t^k$ $\displaystyle\frac1{1+t}=\sum_{j=0}^{\infty}(-1)^jt^j$ The coefficient of $t^n$ in the product is $\displaystyle\sum_{k=1}^{n}\frac{(-1)^{k+1}}k(-1)^{n-k}$ This coefficient is $(-1)^{n+1}H_n$ $\displaystyle\frac{\ln(1+t)}{1+t}=t-\frac32t^2+\frac{11}{6}t^3-\frac{25}{12}t^4+\frac{137}{60}t^5-\frac{49}{20}t^6+\cdots$ $\displaystyle\frac{\ln(1+t)}{1+t}=\sum_{n=1}^{\infty}(-1)^{n+1}H_nt^n$ for $\|t\|<1$	6
(6 marks)		6
Notes
Multiply the standard logarithmic series by the geometric series for $1/(1+t)$ . The coefficient of $t^n$ is $\sum_{k=1}^{n}(-1)^{k+1}(-1)^{n-k}/k=(-1)^{n+1}H_n$ . Substituting $n=1$ to $6$ gives the displayed terms, and the common range is $\|t\|<1$ .
5	$\displaystyle\frac1{1+x^2}=\sum_{r=0}^{\infty}(-1)^rx^{2r}$ for $\|x\|<1$ $\displaystyle\arctan x=\sum_{r=0}^{\infty}\frac{(-1)^rx^{2r+1}}{2r+1}$ $\displaystyle\frac{x}{1+x^2}=\sum_{r=0}^{\infty}(-1)^rx^{2r+1}$ $\displaystyle\arctan x-\frac{x}{1+x^2}=\frac{2x^3}{3}-\frac{4x^5}{5}+\frac{6x^7}{7}-\frac{8x^9}{9}+\cdots$ For $r\ge1$ , the coefficient of $x^{2r+1}$ is $(-1)^r[(2r+1)^{-1}-1]$ $\displaystyle\arctan x-\frac{x}{1+x^2}=\sum_{r=1}^{\infty}(-1)^{r+1}\frac{2r}{2r+1}x^{2r+1}$ The expansion is valid for $\|x\|<1$	7
(7 marks)		7
Notes
Integrating the geometric expansion of $1/(1+x^2)$ gives the series for $\arctan x$ , while multiplication by $x$ gives the series for $x/(1+x^2)$ . Subtraction removes the linear terms and gives coefficient $(-1)^{r+1}2r/(2r+1)$ for $x^{2r+1}$ , starting at $r=1$ . The common range is $\|x\|<1$ .

CP-4.6 · Recognise and use the Maclaurin series for e^x, ln(1+x), sin x, cos x and (1+x)^n, and be aware of the range of values of x for which they are valid (proof not required).

Tier 1 · Easy

Mark scheme for CP-4.6 Tier 1 · Easy
Question	Scheme	Marks
1	$1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+\cdots$ Valid for all real $t$	3
(3 marks)		3
Notes
Use the standard cosine series, which contains even powers with alternating signs: $\cos t=1-t^2/2!+t^4/4!-t^6/6!+\cdots$ . The range of values of $t$ for which the expansion is valid is all real $t$ .
2	$1+0.2+\dfrac{0.2^2}{2!}+\dfrac{0.2^3}{3!}$ $e^{0.2}\approx1.22133$ The estimate is low by about $0.00007$ , or $0.00546\%$ of the stated true value	3
(3 marks)		3
Notes
The four-term estimate is $1+1/5+(1/5)^2/2+(1/5)^3/6=458/375=1.221333\ldots$ , so it is $1.22133$ to five decimal places. Compared with $1.22140$ , it is an underestimate by $1/15000=0.0000666\ldots$ , approximately $0.00007$ or $0.00546\%$ .

Tier 2 · Standard

Mark scheme for CP-4.6 Tier 2 · Standard
Question	Scheme	Marks
1	$1+x+\frac32x^2+\frac52x^3+\cdots$ $\|x\|<\frac12$	4
(4 marks)		4
Notes
In $(1+u)^n$ , take $n=-1/2$ and $u=-2x$ . The first four terms are $1-\frac12u+\frac38u^2-\frac5{16}u^3$ , giving $1+x+\frac32x^2+\frac52x^3$ . The binomial condition $\|u\|<1$ becomes $\|-2x\|<1$ , so $\|x\|<1/2$ .
2	$e^t=1+t+\dfrac{t^2}{2}+\dfrac{t^3}{6}+\cdots$ and $(1-t)^{-1/2}=1+\dfrac{t}{2}+\dfrac{3t^2}{8}+\dfrac{5t^3}{16}+\cdots$ The constant and linear terms are $1+\dfrac32t$ The coefficient of $t^2$ is $\dfrac38+\dfrac12+\dfrac12=\dfrac{11}{8}$ The coefficient of $t^3$ is $\dfrac5{16}+\dfrac38+\dfrac14+\dfrac16=\dfrac{53}{48}$ $\|t\|<1$	5
(5 marks)		5
Notes
Use $e^t=1+t+t^2/2+t^3/6+\cdots$ and $(1-t)^{-1/2}=1+t/2+3t^2/8+5t^3/16+\cdots$ . Multiplication and collection up to and including the term in $t^3$ give coefficients $3/2$ , $11/8$ and $53/48$ . The exponential is valid for every real $t$ , while the non-terminating binomial series requires $\|t\|<1$ .
3	$(1+4u)^{1/2}=1+2u-2u^2+4u^3+\cdots$ $(1-4u)^{1/2}=1-2u-2u^2-4u^3+\cdots$ The constant and even-power terms cancel $f(u)=4u+8u^3+\cdots$ The expansion is valid for $\|u\|<\dfrac14$	5
(5 marks)		5
Notes
Apply the binomial series with power $1/2$ to $1+4u$ and $1-4u$ . Subtracting cancels the constant and every even-power term, leaving $4u+8u^3+\cdots$ . Both component series require $\|4u\|<1$ , so the range is $\|u\|<1/4$ .

Tier 3 · Hard

Mark scheme for CP-4.6 Tier 3 · Hard
Question	Scheme	Marks
1	$2(x+x^3/3+x^5/5+\cdots)$ $-1<x<1$ $\ln(3/2)\approx0.40546$	6
(6 marks)		6
Notes
$\ln(1+x)=x-x^2/2+x^3/3-x^4/4+x^5/5-\cdots$ , while $\ln(1-x)=-x-x^2/2-x^3/3-x^4/4-x^5/5-\cdots$ . Subtracting cancels the even powers and gives the stated series. Both component series are valid together for $-1<x<1$ . To make $(1+x)/(1-x)=3/2$ , use $x=1/5$ . Then $2(1/5+(1/5)^3/3+(1/5)^5/5)=0.405461\ldots$ , so $\ln(3/2)\approx0.40546$ .
2	$\ln(1+t)=t-\dfrac{t^2}{2}+\dfrac{t^3}{3}-\cdots$ $\ln(1-t)=-t-\dfrac{t^2}{2}-\dfrac{t^3}{3}-\cdots$ Coefficient of $t^2$ : $-1$ Coefficient of $t^4$ : $-5/12$ Coefficient of $t^6$ : $-47/180$ $-1<t<1$ $\ln(4/3)\ln(2/3)\approx-0.1166$ ; equivalently, the retained terms give $-7651/65610$	7
(7 marks)		7
Notes
Multiply $t-t^2/2+t^3/3-\cdots$ by $-t-t^2/2-t^3/3-\cdots$ and retain all products up to and including the term in $t^6$ . The odd powers cancel, leaving $-t^2-5t^4/12-47t^6/180+\cdots$ . The intervals $-1<t\le1$ and $-1\le t<1$ intersect in $-1<t<1$ . Setting $t=1/3$ makes the logarithms $\ln(4/3)$ and $\ln(2/3)$ ; the retained terms total $-1/9-5/972-47/131220=-7651/65610$ .
3	$\ln(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}-\dfrac{x^6}{6}+\cdots$ $\sin x=x-\dfrac{x^3}{6}+\dfrac{x^5}{120}+\cdots$ $\ln(1+x)-\sin x=-\dfrac{x^2}{2}+\dfrac{x^3}{2}-\dfrac{x^4}{4}+\dfrac{23x^5}{120}-\dfrac{x^6}{6}+\cdots$ Divide by $x^2$ and use the defined value at $x=0$ $f(x)=-\dfrac12+\dfrac x2-\dfrac{x^2}{4}+\dfrac{23x^3}{120}-\dfrac{x^4}{6}+\cdots$ The expansion is valid for $-1<x\le1$ $f(0.1)\approx-\dfrac12+\dfrac1{20}-\dfrac1{400}+\dfrac{23}{120000}-\dfrac1{60000}=-0.452325$ , so $f(0.1)\approx-0.4523$ to $4$ decimal places	7
(7 marks)		7
Notes
Subtract the standard sine series from the logarithm series through $x^6$ , then divide by $x^2$ . The constant term is $-1/2$ , agreeing with the value defined at zero. This gives $f(x)=-1/2+x/2-x^2/4+23x^3/120-x^4/6+\cdots$ . The logarithm controls the endpoint range, so $-1<x\le1$ . Substitution of $x=0.1$ in the retained terms gives $-0.452325$ , which is $-0.4523$ to $4$ decimal places.
4	$\ln(1+kz)=kz-\dfrac{k^2z^2}{2}+\dfrac{k^3z^3}{3}-\dfrac{k^4z^4}{4}+\cdots$ $\cos z=1-\dfrac{z^2}{2}+\dfrac{z^4}{24}+\cdots$ The coefficient of $z$ is $k$ and the coefficient of $z^2$ is $-k^2/2$ The coefficient of $z^3$ is $k^3/3-k/2$ $k^3/3-k/2=0$ , so $k^2=3/2$ $k=\sqrt{3/2}=\sqrt6/2$ since $k>0$ The coefficient of $z^4$ is $-k^4/4+k^2/4=-3/16$ , so the expansion is $\dfrac{\sqrt6}{2}z-\dfrac34z^2-\dfrac3{16}z^4+\cdots$ The expansion is valid for $-\sqrt{2/3}<z\le\sqrt{2/3}$	8
(8 marks)		8
Notes
Multiply the logarithmic and cosine series through $z^4$ . The cubic coefficient is $k^3/3-k/2$ , so positivity gives $k=\sqrt{3/2}$ . Substitution gives $\ln(1+kz)\cos z=(\sqrt6/2)z-(3/4)z^2-(3/16)z^4+\cdots$ . The logarithmic condition $-1<kz\le1$ gives $-\sqrt{2/3}<z\le\sqrt{2/3}$ .
5	$e^{2w}=1+2w+2w^2+\dfrac43w^3+\dfrac23w^4+\dfrac4{15}w^5+\dfrac4{45}w^6+\cdots$ and $\cos w=1-\dfrac{w^2}{2}+\dfrac{w^4}{24}-\dfrac{w^6}{720}+\cdots$ The given coefficients give $2m-n/2=5$ and $4m/3=4$ $m=3$ $n=2$ The coefficient of $w^4$ is $3(2/3)+2(1/24)=25/12$ The coefficient of $w^5$ is $3(4/15)=4/5$ The coefficient of $w^6$ is $3(4/45)-2/720=19/72$ , so the expansion is $5+6w+5w^2+4w^3+\dfrac{25}{12}w^4+\dfrac45w^5+\dfrac{19}{72}w^6+\cdots$ The expansion is valid for all real $w$	8
(8 marks)		8
Notes
The quadratic and cubic coefficients give $2m-n/2=5$ and $4m/3=4$ , so $m=3$ and $n=2$ . Substitution into the standard exponential and cosine series gives $5+6w+5w^2+4w^3+25w^4/12+4w^5/5+19w^6/72+\cdots$ . Both component series are valid for every real $w$ .