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(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 in (a)(i) above 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,
$\int_{1}^{5} \sqrt{\left(2 x+8 x^{2}\right)} \mathrm{d} x$
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(a)(i) Using the Paschal triangle,
$(1+x)^{7}=1+7 x+21 x^{2}+35 x^{3}+35 x^{4}+21 x^{5}+7 x^{6}+x^{7}$
(ii) $T_{5}=35 ; T_{6}=21 ; T_{7}=7$
$d=T_{6}-T_{5}=21-35=-14$
(b) $\int_{1}^{5} \sqrt{\left(2 x+8 x^{2}\right)} \mathrm{d} x$
\begin{align}
\begin{array}{|l|l|l|l|l|l|}
\hline x & 1 & 2 & 3 & 4 & 5 \\
\hline 2 x & 2 & 4 & 6 & 8 & 10 \\
\hline 8 x^{2} & 8 & 32 & 72 & 128 & 200 \\
\hline 2 x+8 x^{2} & 10 & 36 & 78 & 136 & 210 \\
\hline \sqrt{2 x+8 x^{2}} & 3.162 & 6.0 & 8.832 & 11.662 & 14.491 \\
\hline
\end{array}
\end{align}
\begin{aligned} &y_{1}=3.162, y_{2}=6.0, y_{3}=8.832, y_{4}=11.662, y_{5}=14.491 \\ &y_{1}+y_{5}=3.162+14.491=17.653 \\ &y_{2}+y_{3}+y_{4}=6.0+8.832+11.662=26.494 \\ &\int_{1}^{5} \sqrt{\left(2 x+8 x^{2}\right)} \mathrm{d} x=\frac{1}{2}\left[y_{1}+y_{5}+2\left(y_{2}+y_{3}+y_{4}\right)\right] \\ &=\frac{1}{2}[17.653+2(26.494)] \\ &=\frac{1}{2}[70.641] \\ &=35.3205 \cong 35.32 \end{aligned}
by Diamond (89,036 points)

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