Combinatorics & Binomial

6 methods · 8 questions

★ Signature method unlocked: the technique this chapter rewards most often, with 2 worked questions.

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Many problems hand you an awkward expression whose real identity is a standard series in disguise. Spot the pattern (signs, $1, 2, 3, \dots$ coefficients, a product of $(1 - x^{k})^{- 1}$ factors) and replace it by the closed binomial or geometric form, after which the wanted coefficient is a single formula rather than a long multiplication.

Trigger: "find the coefficient of $x^{n}$ " in something that looks worse than it is, or a sum you suspect equals a clean series.

Instances:
(i) $1 - 2 x + 3 x^{2} - 4 x^{3} + \dots = (1 + x)^{- 2}$ and similar, so a cube becomes $(1 + x)^{- 6}$ and the coefficient comes straight from $(6 - 6)$ .
(ii) split a rational function by partial fractions and expand each piece as a geometric or binomial series in $x$ , then add the coefficients.
(iii) read $(1 - x^{k})^{- 1} = 1 + x^{k} + x^{2 k} + \dots$ as a counting (partition) generating function and count the ways the exponents add to $n$ .
(iv) identify a central binomial coefficient $(r 2 r)$ as the coefficient in $(1 - p)^{- 1/2}$ , then evaluate a series by choosing $p$ .

Linked questions (2)

STEP 2 2008 · Q2

Let

a_{n}

be the coefficient of

x^{n}

in the series expansion, in ascending powers of

x

, of

\frac{1 + x}{( 1 - x ) ^{2} ( 1 + x ^{2} )},

where

∣ x ∣ < 1

. Show, using partial fractions, that either

a_{n} = n + 1

a_{n} = n + 2

according to the value of

n

.

Hence find a decimal approximation, to nine significant figures, for the fraction

\frac{11 000}{8181}

.

[You are not required to justify the accuracy of your approximation.]

STEP 2 2012 · Q1

Write down the general term in the expansion in powers of

x

(1 - x^{6})^{- 2}

Find the coefficient of $x^{24}$ in the expansion in powers of $x$ of $(1 - x^{6})^{- 2} (1 - x^{3})^{- 1} .$ Obtain also, and simplify, formulae for the coefficient of $x^{n}$ in the different cases that arise.
Show that the coefficient of $x^{24}$ in the expansion in powers of $x$ of $(1 - x^{6})^{- 2} (1 - x^{3})^{- 1} (1 - x)^{- 1}$ is $55$ , and find the coefficients of $x^{25}$ and $x^{66}$ .