A) 1/2
B) 1
C) 4/3
D) 3/2
Correct Answer: A
Solution :
\[{{a}_{n}}=\int\limits_{0}^{\pi /2}{(1-\sin t)}{{\,}^{n}}\sin 2tdt\] Let \[1-\sin t=u\Rightarrow -\cos tdt=du\] \[=2\int\limits_{0}^{1}{{{u}^{n}}(1-u)du=2}\left( \int\limits_{0}^{1}{{{u}^{n}}du-}\int\limits_{0}^{1}{{{u}^{n+1}}}du \right)\] \[=2\left( \frac{1}{n+1}-\frac{1}{n+2} \right)\] Hence. \[\frac{{{a}_{n}}}{n}=2\left( \frac{1}{n(n+1)}-\frac{1}{n(n+2)} \right)\] \[\underset{x\to \infty }{\mathop{\lim }}\,\sum\limits_{1}^{n}{\frac{{{a}_{n}}}{n}}=2\left( \sum\limits_{{}}^{{}}{\left( \frac{1}{n}-\frac{1}{n+1} \right)} \right)-\frac{1}{2}\sum\limits_{{}}^{{}}{\left( \frac{1}{n}-\frac{1}{n+2} \right)}\]\[=2\left( \sum\limits_{1}^{n}{\left( \frac{1}{n}-\frac{1}{n+1} \right)} \right)-\sum\limits_{1}^{n}{\left( \frac{1}{n}-\frac{1}{n+2} \right)}\] \[=2(1)-\left[ \left( 1-\frac{1}{3} \right)+\left( \frac{1}{2}-\frac{1}{4} \right)+\left( \frac{1}{3}-\frac{1}{5} \right)+....... \right]\] \[=2-\frac{3}{2}=\frac{1}{2}\]
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