Pure: Sequences, Series and Binomial Depth
As part of A-level Pure Mathematics, sequences, series and binomial at depth covers the binomial series for (1+x)^n when n is not a positive integer, including the validity range this requires, sigma-notation sums that need splitting into standard results, recurrence relations that generate arithmetic, geometric, periodic or other sequences, and the sum to infinity of a convergent geometric series used in both proofs and modelling contexts. The marks reward stating the validity range for a binomial expansion (routinely dropped), using partial fractions before expanding a rational function binomially, and identifying what kind of sequence a recurrence generates before applying a formula to it.
Before you start
Make sure you're comfortable with these topics first:
Method
- For (1+x)^n with n a negative or non-integer rational number, use the binomial series (1+x)^n = 1 + nx + [n(n-1)/2!]x^2 + [n(n-1)(n-2)/3!]x^3 + ..., an infinite series valid only for |x| < 1 - always state this range as part of the answer.
- For (a+bx)^n where a is not 1, write it as a^n(1+(b/a)x)^n first, expand the bracket using the standard series in (b/a)x, then multiply every term by a^n; the validity range becomes |x| < |a/b|.
- If the expression to expand is a rational function (for example from a partial-fractions decomposition), expand each partial fraction separately as its own binomial series, then collect like powers of x; the overall validity range is the more restrictive (smaller) of the individual ranges.
- For sigma-notation sums, split a sum of several terms into separate standard sums (such as a sum of r^2 minus a multiple of a sum of r plus a constant sum), and substitute the standard formulae [n(n+1)/2 for sum of r, n(n+1)(2n+1)/6 for sum of r^2] before simplifying algebraically.
- To identify what kind of sequence a recurrence relation generates, compute the first few terms by hand: a constant difference means arithmetic (use u_n = a+(n-1)d), a constant ratio means geometric (use u_n = ar^(n-1), checking |r| < 1 before assuming a sum to infinity exists), and terms that repeat means periodic (identify the period, then use it to jump straight to any required term).
- For a geometric series sum to infinity, first justify convergence by checking |r| < 1 (if the question does not already state it), then apply S_infinity = a/(1-r).
- When a 'show that' question gives a recurrence and a claim about a term or a sum, work directly from the recurrence (substituting the given u_n and simplifying algebraically) rather than guessing and checking, since this generalises to any term the question might later ask about.
Worked example
Find the binomial expansion of 1/sqrt(4-3x), in ascending powers of x, up to and including the term in x^3, simplifying each coefficient, and state the range of values of x for which the expansion is valid.
- Write 4 - 3x = 4(1 - (3/4)x), so 1/sqrt(4-3x) = 1/(2sqrt(1-(3/4)x)) = (1/2)(1-(3/4)x)^(-1/2).
- Use the binomial series (1+u)^n = 1 + nu + [n(n-1)/2!]u^2 + [n(n-1)(n-2)/3!]u^3 + ... with n = -1/2 and u = -(3/4)x.
- Find each coefficient: the u term has coefficient -1/2; the u^2 term has coefficient [(-1/2)(-3/2)]/2 = 3/8; the u^3 term has coefficient [(-1/2)(-3/2)(-5/2)]/6 = -5/16.
- Substitute u = -(3/4)x into each term: -(1/2)u = (3/8)x; (3/8)u^2 = (27/128)x^2; -(5/16)u^3 = (135/1024)x^3.
- Multiply every term by the factor of 1/2 from step 1: 1/sqrt(4-3x) = 1/2 + (3/16)x + (27/256)x^2 + (135/2048)x^3 + ...
- State the validity range: since the series requires |u| < 1, i.e. |(3/4)x| < 1, the expansion is valid for |x| < 4/3.
Practice questions
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Q1State the validity condition (range of x) for the binomial expansion of (1+x)^n to be used, when n is not a positive integer.Show answer
Answer: |x| < 1 (i.e. -1 < x < 1).
Q2A sequence is defined by u_1 = 3, u_(n+1) = 2u_n - 1 for n >= 1. Find u_2, u_3 and u_4, and state whether the sequence is arithmetic, geometric or neither.Show answer
Answer: u_2 = 5, u_3 = 9, u_4 = 17. Neither: the differences (2, 4, 8) are not constant, so it is not arithmetic, and the ratios (5/3, 9/5, 17/9) are not constant, so it is not geometric.
Q3A sequence is defined by u_1 = 4, u_(n+1) = 1/(1-u_n) for n >= 1. Find u_2, u_3 and u_4, and hence write down the value of u_100.Show answer
Answer: u_2 = -1/3, u_3 = 3/4, u_4 = 4. Since u_4 = u_1, the sequence is periodic with period 3; since 100 = 3(33) + 1, u_100 occupies the same position in the cycle as u_1, so u_100 = 4.
Q4Find sum from r=1 to 15 of (3r-2).Show answer
Answer: 330. (Split as 3 sum(r) - sum(2) = 3(15)(16)/2 - 2(15) = 360 - 30.)
Q5Find the coefficient of x^3 in the binomial expansion of (1-2x)^(-3), in ascending powers of x.Show answer
Answer: 80. (The general term for k=3, n=-3 gives coefficient [(-3)(-4)(-5)]/3! = -10, and (-2x)^3 = -8x^3, so the term is -10 x (-8x^3) = 80x^3.)
Q6Explain why the binomial expansion of (2+5x)^(-1) must first be written as (1/2)(1+(5/2)x)^(-1) before the standard series 1-u+u^2-u^3+... can be applied, rather than expanding (2+5x)^(-1) directly.Show answer
Answer: The standard binomial series (1+u)^n = 1+nu+... only applies to a bracket with a leading 1. Since (2+5x) does not start with 1, a factor of 2 must first be taken out to write it as 2(1+(5/2)x), with 2^(-1) taken outside; skipping this step and applying the series to (2+5x) directly would use the wrong 'u' and give both an incorrect expansion and an incorrect validity range (which should be based on |(5/2)x| < 1, not |5x| < 1).
Q7A ball is dropped from a height of 4 m and bounces repeatedly. After each bounce it rises to 0.7 of the height it fell from. Find the total distance travelled by the ball by the time it comes to rest.Show answer
Answer: 68/3 m, approximately 22.7 m (3 s.f.). Total = initial fall (4) + 2 x (sum to infinity of the bounce heights) = 4 + 2 x [4(0.7)/(1-0.7)] = 4 + 56/3.
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
f(x) = (5-x)/((1-x)(2+x)), where |x| < 1. (a) Express f(x) in partial fractions. (4) (b) Hence find the binomial expansion of f(x) in ascending powers of x, up to and including the term in x^2, simplifying each coefficient. (5)
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Prove that the recurring decimal 0.454545... (repeating) can be written as the fraction 5/11, by expressing it as the sum of an infinite geometric series.
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Show that sum from r=1 to n of (6r^2 - 2) = n(2n^2 + 3n - 1). (4)
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