Edexcel A-level Maths coverage

Sequences and series

Section 4
6 spec leafs

Notes and three levels of exam-style practice for each registered specification leaf in this section.

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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 nn, (a+bx)n=r=0n(nr)anr(bx)r(a+bx)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}(bx)^r, where (nr)=n!/[r!(nr)!]\binom{n}{r}=n!/[r!(n-r)!], n!=n(n1)1n!=n(n-1)\cdots1 and 0!=10!=1; Pascal's relation is (nr)=(n1r1)+(n1r)\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r} for 1rn11\le r\le n-1.
  • The same coefficients appear in binomial probabilities: P(X=r)=(nr)pr(1p)nrP(X=r)=\binom{n}{r}p^r(1-p)^{n-r} for XB(n,p)X\sim B(n,p).
  • For rational nn, write the expression as an(1+u)na^n(1+u)^n and use 1+nu+n(n1)u2/2!+1+nu+n(n-1)u^2/2!+\cdots, valid for u<1|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 bb from terms involving (bx)r(bx)^r.

Tier 1 · Easy

2 marks
ORIGINAL

Find the coefficient of x2x^2 in the expansion of (1+4x)5(1+4x)^5.

Tier 2 · Standard

4 marks
ORIGINAL

For (12x)1/2(1-2x)^{-1/2}, obtain the constant, xx and x2x^2 terms. Also give the interval of xx on which this series is valid.

Tier 3 · Hard

5 marks
ORIGINAL

Use the first three terms of a binomial expansion of (64+x)1/3(64+x)^{1/3} to estimate 653\sqrt[3]{65}. Give the estimate to 55 decimal places and justify that the expansion is valid at the value of xx used.

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 xnx_n directly from nn, whereas a recurrence relation defines each term from one or more preceding terms and needs an initial value.
  • A sequence is increasing when xn+1>xnx_{n+1}>x_n and decreasing when xn+1<xnx_{n+1}<x_n throughout the stated range of nn.
  • 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 ff repeatedly to the initial term instead of to the latest term.

Tier 1 · Easy

2 marks
ORIGINAL

The sequence (xn)(x_n) is defined by xn=73nx_n=7-3n for n1n\ge1. Write down its first three terms and state whether it is increasing or decreasing.

Tier 2 · Standard

3 marks
ORIGINAL

A sequence is defined by x1=1x_1=1 and xn+1=5xnx_{n+1}=5-x_n. Find x2,x3,x4,x5x_2,x_3,x_4,x_5 and state its period.

Tier 3 · Hard

5 marks
ORIGINAL

A sequence is defined by x1=2x_1=2 and xn+1=(xn+6)/2x_{n+1}=(x_n+6)/2. Let yn=6xny_n=6-x_n. Show that (yn)(y_n) is geometric, find a formula for xnx_n, and hence show that (xn)(x_n) is increasing.

4.3

Understand and use sigma notation for sums of series.

  • The notation r=abf(r)\sum_{r=a}^{b}f(r) means add the values of f(r)f(r) for each integer rr from aa to bb 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, r=1n(ar+b)=ar=1nr+br=1n1\sum_{r=1}^{n}(ar+b)=a\sum_{r=1}^{n}r+b\sum_{r=1}^{n}1, with r=1n1=n\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 11.

Tier 1 · Easy

2 marks
ORIGINAL

Evaluate r=14(2r+1)\sum_{r=1}^{4}(2r+1).

Tier 2 · Standard

2 marks
ORIGINAL

Write the series 5+9+13++415+9+13+\cdots+41 using sigma notation.

Tier 3 · Hard

5 marks
ORIGINAL

Given that r=1n(4r1)=210\sum_{r=1}^{n}(4r-1)=210, find the positive integer nn. Show all stages of your working.

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 dd and nth term un=a+(n1)du_n=a+(n-1)d, where aa is the first term.
  • The sum of its first nn terms is Sn=n[2a+(n1)d]/2S_n=n[2a+(n-1)d]/2, equivalently Sn=n(a+l)/2S_n=n(a+l)/2 when the last term ll is known.
  • Pairing the first and last terms, then the second and penultimate terms, gives nn equal pairs of total a+la+l across two copies of the series, proving Sn=n(a+l)/2S_n=n(a+l)/2.
  • The term number nn must be a positive integer. A common error is to use a+nda+nd for the nth term, shifting every term by one difference.

Tier 1 · Easy

2 marks
ORIGINAL

Find the 1818th term of the arithmetic sequence 11,17,23,11,17,23,\ldots.

Tier 2 · Standard

3 marks
ORIGINAL

An arithmetic series has first term 77, last term 151151 and common difference 88. Find the number of terms and the sum of the series.

Tier 3 · Hard

5 marks
ORIGINAL

An arithmetic sequence has first term aa and common difference dd. Its 88th term is 3131 and the sum of its first 2020 terms is 730730. Find aa and dd, and find the least value of nn for which the nth term exceeds 100100.

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 rr and nth term un=arn1u_n=ar^{n-1}, where aa is the first term.
  • For r1r\ne1, subtracting rSnrS_n from SnS_n proves Sn=a(1rn)/(1r)S_n=a(1-r^n)/(1-r); an equivalent form may be more convenient when r>1r>1.
  • A geometric series has a finite sum to infinity only when r<1|r|<1, in which case S=a/(1r)S_\infty=a/(1-r); for a finite-sum threshold, isolate rnr^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<1r<1 instead of r<1|r|<1, which would wrongly accept ratios below 1-1.

Tier 1 · Easy

2 marks
ORIGINAL

Find the 88th term of the geometric sequence 3,6,12,3,6,12,\ldots.

Tier 2 · Standard

3 marks
ORIGINAL

Find the exact sum to infinity of 123+0.7512-3+0.75-\cdots.

Tier 3 · Hard

5 marks
ORIGINAL

A geometric series has first term 800800 and common ratio 0.90.9. Find the least value of nn for which the sum of the first nn terms exceeds 70007000.

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 u1u_1 or u0u_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

2 marks
ORIGINAL

Mina saves £25\pounds 25 in the first month and increases the amount saved by £5\pounds 5 each month. Find the total she saves in the first 66 months.

Tier 2 · Standard

4 marks
ORIGINAL

A ball is dropped from a height of 1212 m. After each impact it rebounds to 65%65\% of its previous maximum height. Find the total vertical distance travelled before it comes to rest, giving your answer to 33 significant figures.

Tier 3 · Hard

6 marks
ORIGINAL

An account initially contains £5000\pounds 5000. At the end of each year, after interest of 3%3\% has been added, £400\pounds 400 is withdrawn. The model is B0=5000B_0=5000 and Bn+1=1.03Bn400B_{n+1}=1.03B_n-400. Derive a formula for BnB_n and find the first value of nn for which the model predicts a negative balance. State one limitation of the model.