WAEC 2016 · Paper 2 · Q10✱✱

(a)
(i) Write down the expansion of $(1 + x)^7$ in ascending powers of $x$ . (ii) If the coefficients of the fifth, sixth and seventh terms in the expansion form a linear sequence (A.P.), find the common difference of the A.P.
(b)
Using the trapezium rule with ordinates at $1, 2, 3, 4$ and $5$ , calculate, correct to two decimal places, $\displaystyle\int_1^5 \sqrt{2x + 8x^{-2}}\,dx$ .

Worked solution (try it first)

(a)(i)

The coefficients are row 7 of Pascal's triangle: $(1 + x)^7 = 1 + 7x + 21x^2 + 35x^3 + 35x^4 + 21x^5 + 7x^6 + x^7$ .

(ii)

The fifth, sixth and seventh terms are $35x^4$ , $21x^5$ and $7x^6$ , so the coefficients are $35, 21, 7$ .
They go down by 14 each time: $21 - 35 = -14$ and $7 - 21 = -14$ .
The common difference is $-14$ .

(b)

$h = 1$ .
$y = \sqrt{2x + 8x^{-2}}$ gives $3.1623,\ 2.4495,\ 2.6247,\ 2.9155,\ 3.2125$ at $x = 1, \ldots, 5$ .
$\frac12[(3.1623 + 3.2125) + 2(2.4495 + 2.6247 + 2.9155)] = \frac12[6.3748 + 15.9794]$
$= 11.1771$ , about $11.18$ .

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