(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,
\(\int_{1}^{5} \sqrt{(2x + 8x^{2})} \mathrm {d} x\).
(a)(i) Using the Paschal triangle,
\((1 + x)^{7} = 1 + 7x + 21x^{2} + 35x^{3} + 35x^{4} + 21x^{5} + 7x^{6} + x^{7}\)
(ii) \(T_{5} = 35 ; T_{6} = 21 ; T_{7} = 7\)
\(d = T_{6} - T_{5} = 21 - 35 = -14\)
(b) \(\int_{1}^{5} \sqrt{(2x + 8x^{2})} \mathrm {d} x\)
| x | 1 | 2 | 3 | 4 | 5 |
| \(2x\) | 2 | 4 | 6 | 8 | 10 |
| \(8x^{2}\) | 8 | 32 | 72 | 128 | 200 |
| \(2x + 8x^{2}\) | 10 | 36 | 78 | 136 | 210 |
| \(\sqrt{2x + 8x^{2}}\) | 3.162 | 6.0 | 8.832 | 11.662 | 14.491 |
h = 1
\(y_{1} = 3.162, y_{2} = 6.0 , y_{3} = 8.832 , y_{4} = 11.662 , y_{5} = 14.491\)
\(y_{1} + y_{5} = 3.162 + 14.491 = 17.653\)
\(y_{2} + y_{3} + y_{4} = 6.0 + 8.832 + 11.662 = 26.494\)
\(\int_{1}^{5} \sqrt{(2x + 8x^{2})} \mathrm {d} x = \frac{1}{2}[y_{1} + y_{5} + 2(y_{2} + y_{3} + y_{4})]\)
= \(\frac{1}{2} [17.653 + 2(26.494)]\)
= \(\frac{1}{2} [70.641]\)
= \(35.3205 \approxeq 35.32\)
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}