Q46 / 90 · KVPY 2014 (SB/SX) · Mathematics
Let $I_{n}=\int_{0}^{\pi / 2} x^{n} \cos x d x$, where $n$ is a non negative integer Then $\sum_{n=2}^{\infty}\left(\frac{I_{n}}{n !}+\frac{I_{n-2}}{(n-2) !}\right)$ equal
- Correct answerA$e^{\pi / 2}-1-\frac{\pi}{2}$
- B$e^{\pi / 2}-1$
- C$e^{\pi / 2}-\frac{\pi}{2}$
- D$\mathrm{e}^{\pi / 2}$
Solution
$$
\begin{array}{l}
\text { Sol- } I_{n}=\int_{0}^{\pi / 2} x^{n} \cos x d x \\
\Rightarrow I_{n}+n(n-1) I_{n-2}=\left(\frac{\pi}{2}\right)^{n}
\end{array}
$$
Also $\sum_{n=2}^{\infty}\left(\frac{1_{n}}{n !}+\frac{I_{n-2}}{(n-2) !}\right)=\sum_{n=2}^{\infty}\left(\frac{I_{n}+(n-2) ! I_{n-2}}{n !}\right)=\sum_{n=2}^{\infty}\left(\frac{(\pi / 2)^{n}}{n !}\right)=e^{\pi / 2}-1-\frac{\pi}{2}$
Hence, the correct answer is option: 'A'.