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