4 Sequences and series — printable coverage pack

4.1 · Understand and use the binomial expansion of (a + bx)ⁿ for positive integer n; the notations n! and nCr; link to binomial probabilities; extend to any rational n, including for approximation, valid for |bx/a| < 1.

For a positive integer $n$ , $(a+bx)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}(bx)^r$ , where $\binom{n}{r}=n!/[r!(n-r)!]$ , $n!=n(n-1)\cdots1$ and $0!=1$ ; Pascal's relation is $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$ for $1\le r\le n-1$ .
The same coefficients appear in binomial probabilities: $P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}$ for $X\sim B(n,p)$ .
For rational $n$ , write the expression as $a^n(1+u)^n$ and use $1+nu+n(n-1)u^2/2!+\cdots$ , valid for $|u|<1$ .
For an approximation, choose a nearby convenient value and retain the requested number of terms. A common error is to omit powers of the coefficient $b$ from terms involving $(bx)^r$ .

Tier 1 · Easy

1. Find the coefficient of $x^2$ in the expansion of $(1+4x)^5$ .[2 marks]
Answer
- $160$
Method: The $x^2$ term is obtained with $r=2$ : $\binom{5}{2}(4x)^2=10\times16x^2=160x^2$ . Hence the coefficient is $160$ .

Tier 2 · Standard

1. For $(1-2x)^{-1/2}$ , obtain the constant, $x$ and $x^2$ terms. Also give the interval of $x$ on which this series is valid.[4 marks]
Answer
- $1+x+\dfrac{3}{2}x^2$
- $|x|<\dfrac{1}{2}$
Method: Use $(1+u)^n=1+nu+n(n-1)u^2/2+\cdots$ with $n=-1/2$ and $u=-2x$ . This gives $1+(-1/2)(-2x)+[(-1/2)(-3/2)/2](-2x)^2=1+x+3x^2/2$ . Validity requires $|-2x|<1$ , so $|x|<1/2$ .

Tier 3 · Hard

1. Use the first three terms of a binomial expansion of $(64+x)^{1/3}$ to estimate $\sqrt[3]{65}$ . Give the estimate to $5$ decimal places and justify that the expansion is valid at the value of $x$ used.[5 marks]
Answer
- $\sqrt[3]{65}\approx4.02072$
- Valid because $|1/64|<1$
Method: Write $(64+x)^{1/3}=4(1+x/64)^{1/3}$ . Using $n=1/3$ , the first three terms give $4[1+x/192-x^2/36864]=4+x/48-x^2/9216$ . Set $x=1$ to estimate $\sqrt[3]{65}$ : $4+1/48-1/9216=4.0207248\ldots$ , so the estimate is $4.02072$ . The expansion requires $|x/64|<1$ , which holds when $x=1$ .

4.2 · Work with sequences including those given by a formula for the nth term and those generated by a simple relation of the form xₙ₊₁ = f(xₙ); increasing sequences; decreasing sequences; periodic sequences.

An explicit rule gives $x_n$ directly from $n$ , whereas a recurrence relation defines each term from one or more preceding terms and needs an initial value.
A sequence is increasing when $x_{n+1}>x_n$ and decreasing when $x_{n+1}<x_n$ throughout the stated range of $n$ .
A sequence is periodic if its terms repeat after a fixed positive number of steps; the least such number is its period.
When using a recurrence, retain sufficient accuracy between steps and check any invariant interval. A common error is to apply $f$ repeatedly to the initial term instead of to the latest term.

Tier 1 · Easy

1. The sequence $(x_n)$ is defined by $x_n=7-3n$ for $n\ge1$ . Write down its first three terms and state whether it is increasing or decreasing.[2 marks]
Answer
- $4,1,-2$
- Decreasing
Method: Substitute $n=1,2,3$ to obtain $4,1,-2$ . Also $x_{n+1}-x_n=-3<0$ , so the sequence is decreasing.

Tier 2 · Standard

1. A sequence is defined by $x_1=1$ and $x_{n+1}=5-x_n$ . Find $x_2,x_3,x_4,x_5$ and state its period.[3 marks]
Answer
- $x_2=4$ , $x_3=1$ , $x_4=4$ , $x_5=1$
- Period $2$
Method: Apply the recurrence to the most recent term: $x_2=5-1=4$ , $x_3=5-4=1$ , and the pair $1,4$ then repeats. Thus $x_4=4$ , $x_5=1$ , and the least period is $2$ .

Tier 3 · Hard

1. A sequence is defined by $x_1=2$ and $x_{n+1}=(x_n+6)/2$ . Let $y_n=6-x_n$ . Show that $(y_n)$ is geometric, find a formula for $x_n$ , and hence show that $(x_n)$ is increasing.[5 marks]
Answer
- $y_n=4\left(\dfrac{1}{2}\right)^{n-1}$
- $x_n=6-4\left(\dfrac{1}{2}\right)^{n-1}$
- $(x_n)$ is increasing
Method: Using the recurrence, $y_{n+1}=6-x_{n+1}=6-(x_n+6)/2=(6-x_n)/2=y_n/2$ . Also $y_1=6-2=4$ , so $(y_n)$ is geometric with ratio $1/2$ and $y_n=4(1/2)^{n-1}$ . Therefore $x_n=6-4(1/2)^{n-1}$ . Since $(1/2)^{n-1}$ is positive and decreases as $n$ increases, subtracting it from $6$ shows that $(x_n)$ is increasing.

4.3 · Understand and use sigma notation for sums of series.

The notation $\sum_{r=a}^{b}f(r)$ means add the values of $f(r)$ for each integer $r$ from $a$ to $b$ inclusive.
To write a series in sigma notation, identify a formula for its general term together with the correct first and last index values.
Use linearity to split sums: for example, $\sum_{r=1}^{n}(ar+b)=a\sum_{r=1}^{n}r+b\sum_{r=1}^{n}1$ , with $\sum_{r=1}^{n}1=n$ .
Check the endpoint terms after changing an index. A common error is to treat the upper limit as the number of terms when the lower limit is not $1$ .

Tier 1 · Easy

1. Evaluate $\sum_{r=1}^{4}(2r+1)$ .[2 marks]
Answer
- $24$
Method: Substitute $r=1,2,3,4$ : the sum is $3+5+7+9=24$ .

Tier 2 · Standard

1. Write the series $5+9+13+\cdots+41$ using sigma notation.[2 marks]
Answer
- $\sum_{r=1}^{10}(4r+1)$
Method: The terms have common difference $4$ , and $4r+1$ gives $5$ when $r=1$ . Solving $4r+1=41$ gives $r=10$ , so the series is $\sum_{r=1}^{10}(4r+1)$ .

Tier 3 · Hard

1. Given that $\sum_{r=1}^{n}(4r-1)=210$ , find the positive integer $n$ . Show all stages of your working.[5 marks]
Answer
- $n=10$
Method: By linearity, $\sum_{r=1}^{n}(4r-1)=4[n(n+1)/2]-n=2n^2+n$ . Hence $2n^2+n=210$ , so $2n^2+n-210=0$ . Factorising gives $(2n+21)(n-10)=0$ . Since $n$ is positive, $n=10$ .

4.4 · Understand and work with arithmetic sequences and series, including the formulae for nth term and the sum to n terms.

An arithmetic sequence has constant difference $d$ and nth term $u_n=a+(n-1)d$ , where $a$ is the first term.
The sum of its first $n$ terms is $S_n=n[2a+(n-1)d]/2$ , equivalently $S_n=n(a+l)/2$ when the last term $l$ is known.
Pairing the first and last terms, then the second and penultimate terms, gives $n$ equal pairs of total $a+l$ across two copies of the series, proving $S_n=n(a+l)/2$ .
The term number $n$ must be a positive integer. A common error is to use $a+nd$ for the nth term, shifting every term by one difference.

Tier 1 · Easy

1. Find the $18$ th term of the arithmetic sequence $11,17,23,\ldots$ .[2 marks]
Answer
- $113$
Method: Here $a=11$ and $d=6$ . Thus $u_{18}=11+(18-1)6=11+102=113$ .

Tier 2 · Standard

1. An arithmetic series has first term $7$ , last term $151$ and common difference $8$ . Find the number of terms and the sum of the series.[3 marks]
Answer
- $19$ terms
- $1501$
Method: Use $151=7+(n-1)8$ , giving $144=8(n-1)$ and $n=19$ . Then $S_{19}=19(7+151)/2=19\times79=1501$ .

Tier 3 · Hard

1. An arithmetic sequence has first term $a$ and common difference $d$ . Its $8$ th term is $31$ and the sum of its first $20$ terms is $730$ . Find $a$ and $d$ , and find the least value of $n$ for which the nth term exceeds $100$ .[5 marks]
Answer
- $a=\dfrac{78}{5}$
- $d=\dfrac{11}{5}$
- $n=40$
Method: The term condition gives $a+7d=31$ . The sum condition gives $10(2a+19d)=730$ , so $2a+19d=73$ . Doubling the first equation and subtracting gives $5d=11$ , hence $d=11/5$ and $a=31-77/5=78/5$ . Therefore $u_n=(11n+67)/5$ . Solving $(11n+67)/5>100$ gives $n>433/11=39.36\ldots$ , so the least integer is $40$ .

4.5 · Understand and work with geometric sequences and series, including the formulae for the nth term and the sum of a finite geometric series; the sum to infinity of a convergent geometric series, including the use of |r| < 1; modulus notation.

A geometric sequence has constant ratio $r$ and nth term $u_n=ar^{n-1}$ , where $a$ is the first term.
For $r\ne1$ , subtracting $rS_n$ from $S_n$ proves $S_n=a(1-r^n)/(1-r)$ ; an equivalent form may be more convenient when $r>1$ .
A geometric series has a finite sum to infinity only when $|r|<1$ , in which case $S_\infty=a/(1-r)$ ; for a finite-sum threshold, isolate $r^n$ and use logarithms.
A negative ratio makes term signs alternate, but convergence still depends on its modulus. A common error is to use $r<1$ instead of $|r|<1$ , which would wrongly accept ratios below $-1$ .

Tier 1 · Easy

1. Find the $8$ th term of the geometric sequence $3,6,12,\ldots$ .[2 marks]
Answer
- $384$
Method: The first term is $a=3$ and the common ratio is $r=2$ . Hence $u_8=3\times2^7=384$ .

Tier 2 · Standard

1. Find the exact sum to infinity of $12-3+0.75-\cdots$ .[3 marks]
Answer
- $\dfrac{48}{5}$
Method: The common ratio is $r=-3/12=-1/4$ , and $|r|=1/4<1$ , so the series converges. Its sum is $S_\infty=12/[1-(-1/4)]=12/(5/4)=48/5$ .

Tier 3 · Hard

1. A geometric series has first term $800$ and common ratio $0.9$ . Find the least value of $n$ for which the sum of the first $n$ terms exceeds $7000$ .[5 marks]
Answer
- $n=20$
Method: The finite sum is $S_n=800(1-0.9^n)/(1-0.9)=8000(1-0.9^n)$ . The condition $S_n>7000$ gives $0.9^n<0.125$ . Taking logarithms and accounting for $\ln(0.9)<0$ gives $n>\ln(0.125)/\ln(0.9)=19.73\ldots$ . Therefore the least integer is $n=20$ .

4.6 · Use sequences and series in modelling.

A constant additive change suggests an arithmetic model, while a constant multiplier or percentage change suggests a geometric model.
State what the term number represents and whether the initial value is $u_1$ or $u_0$ before forming a term or sum.
Repeated deposits, withdrawals or other fixed adjustments can be represented by a recurrence and often rewritten using a finite geometric sum.
Interpret results within the context and identify assumptions such as fixed rates or indefinite continuation. A common error is to use an infinite sum when the process has not converged or has a finite stopping point.

Tier 1 · Easy

1. Mina saves $\pounds 25$ in the first month and increases the amount saved by $\pounds 5$ each month. Find the total she saves in the first $6$ months.[2 marks]
Answer
- $\pounds 225$
Method: The monthly amounts form an arithmetic sequence with $a=25$ , $d=5$ and sixth term $25+5(5)=50$ . Therefore $S_6=6(25+50)/2=225$ .

Tier 2 · Standard

1. A ball is dropped from a height of $12$ m. After each impact it rebounds to $65\%$ of its previous maximum height. Find the total vertical distance travelled before it comes to rest, giving your answer to $3$ significant figures.[4 marks]
Answer
- $56.6$ m
Method: The initial downward distance is $12$ m. The rebound heights form a geometric series with first term $12(0.65)=7.8$ and ratio $0.65$ . Each rebound height is travelled once upwards and once downwards, so the total is $12+2[7.8/(1-0.65)]=56.5714\ldots$ m. To $3$ significant figures this is $56.6$ m.

Tier 3 · Hard

1. An account initially contains $\pounds 5000$ . At the end of each year, after interest of $3\%$ has been added, $\pounds 400$ is withdrawn. The model is $B_0=5000$ and $B_{n+1}=1.03B_n-400$ . Derive a formula for $B_n$ and find the first value of $n$ for which the model predicts a negative balance. State one limitation of the model.[6 marks]
Answer
- $B_n=5000(1.03)^n-400\dfrac{(1.03)^n-1}{0.03}$
- $n=16$
- For example, the model assumes the interest rate and withdrawal remain fixed and permits an unrealistic negative balance.
Method: After $n$ years, the initial balance has grown to $5000(1.03)^n$ . The withdrawals accumulate as $400[1+1.03+\cdots+1.03^{n-1}]=400[(1.03)^n-1]/0.03$ , giving the stated formula. Rearranging, $B_n=40000/3-(25000/3)(1.03)^n$ . A negative balance requires $(1.03)^n>1.6$ , so $n>\ln(1.6)/\ln(1.03)=15.90\ldots$ . The first integer is $n=16$ . In reality the account provider would not continue the same process into a negative balance, and the rate may change.