(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...
Question 1 Report
(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,
\(\int_{1}^{5} \sqrt{(2x + 8x^{2})} \mathrm {d} x\).
Answer Details
(a)(i) By the binomial theorem:
\[(1 + x)^7 = 1 + 7x + 21x^2 + 35x^3 + 35x^4 + 21x^5 + 7x^6 + x^7\](ii) The 5th, 6th and 7th coefficients are \(35,\ 21,\ 7\). They form an A.P. with common difference:
\[d = 21 - 35 = -14 \quad(\text{and } 7 - 21 = -14)\](b) Let \(f(x) = \sqrt{2x + 8x^2}\), with \(h = 1\). Compute the ordinates:
- \(f(1) = \sqrt{10} = 3.1623\)
- \(f(2) = \sqrt{36} = 6.0000\)
- \(f(3) = \sqrt{78} = 8.8318\)
- \(f(4) = \sqrt{136} = 11.6619\)
- \(f(5) = \sqrt{210} = 14.4914\)
Trapezium rule:
\[\int_1^5 f\,dx \approx \tfrac{h}{2}\big[(f_0 + f_4) + 2(f_1 + f_2 + f_3)\big]\] \[= \tfrac{1}{2}\big[(3.1623 + 14.4914) + 2(6 + 8.8318 + 11.6619)\big]\] \[= \tfrac{1}{2}\big[17.6537 + 52.9874\big] = \tfrac{1}{2}(70.6411) = 35.32\]So the integral is approximately \(35.32\) (to 2 decimal places).
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