WAEC 2022 · Paper 2 · Q11

Sequences, series & binomial expansion

(a)
Find the binomial expansion of $(1 + 2x)^7$ and $(1 - 2x)^7$ .
Show the answer
$(1 \pm 2x)^7 = 1 \pm 14x + 84x^2 \pm 280x^3 + 560x^4 \pm 672x^5 + 448x^6 \pm 128x^7$
(b)
Using the result in (a), find, correct to three decimal places, the value of $(1.2)^7 - (0.8)^7$ .

Worked solution (try it first)

(a)

Each term of $(1 + 2x)^7$ is $\binom7r(2x)^r$ : $1 + 14x + 84x^2 + 280x^3 + 560x^4 + 672x^5 + 448x^6 + 128x^7$ .
$(1 - 2x)^7$ has the same terms with the odd powers of $x$ negative: $1 - 14x + 84x^2 - 280x^3 + 560x^4 - 672x^5 + 448x^6 - 128x^7$ .

(b)

$(1.2)^7 - (0.8)^7$ is $(1 + 2x)^7 - (1 - 2x)^7$ with $x = 0.1$ .
Subtracting, the even powers cancel and the odd ones double: $28x + 560x^3 + 1344x^5 + 256x^7$ .
With $x = 0.1$ : $2.8 + 0.56 + 0.01344 + 0.0000256 = 3.3734656$ , so the value is $3.373$ to three decimal places.

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