Binomial expansion (positive integer n)

Series

A-level Maths (9MA0)

ORIGINAL

Use the nCr coefficients and raise the whole of the second term — sign included — to each power.

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Question

(a) Find the first four terms, in ascending powers of

x

, of the binomial expansion of

(2 - 3x)^5

, giving each term in its simplest form. (b) Hence find the coefficient of

x^3

in the expansion of

(1 + x)(2 - 3x)^5

Every step worked, with the reasoning.

1
$(a + b)^5 = a^5 + \binom{5}{1}a^4 b + \binom{5}{2}a^3 b^2 + \binom{5}{3}a^2 b^3 + \dots$ with $a = 2$ and $b = -3x$
For a positive integer power use the binomial expansion with ${}^nC_r$ coefficients; keep the minus sign inside $b$ .
2
$= 2^5 + 5(2^4)(-3x) + 10(2^3)(-3x)^2 + 10(2^2)(-3x)^3 + \dots$
Substitute, using $\binom{5}{1} = 5$ , $\binom{5}{2} = 10$ and $\binom{5}{3} = 10$ .
3
$= 32 + 5(16)(-3x) + 10(8)(9x^2) + 10(4)(-27x^3) + \dots$
Evaluate the powers: $(-3x)^2 = 9x^2$ and $(-3x)^3 = -27x^3$ .
4
$= 32 - 240x + 720x^2 - 1080x^3 + \dots$
Multiply out each coefficient.
5
In $(1 + x)(32 - 240x + 720x^2 - 1080x^3 + \dots)$ the $x^3$ terms are $1 \times (-1080x^3)$ and $x \times 720x^2$ .
An $x^3$ term comes from $1$ times the $x^3$ term, or from $x$ times the $x^2$ term.
6
Coefficient of $x^3 = -1080 + 720 = -360$
Add the two contributions.

Answer: (a) $32 - 240x + 720x^2 - 1080x^3$ (b) $-360$