Q71 / 120 · VITEEE 2012 · Mathematics
The sum of the series
$$
1+\frac{1^{2}+2^{2}}{2 !}+\frac{1^{2}+2^{2}+3^{2}}{3 !}+\frac{1^{2}+2^{2}+3^{2}+4^{2}}{4 !}+\ldots
$$
is
- A$3 \mathrm{e}$
- Correct answerB$\frac{17}{6} \mathrm{e}$
- C$\frac{13}{6} \mathrm{e}$
- D$\frac{19}{6} \mathrm{e}$
Solution
$$
\begin{array}{l}
T_{n}=\frac{1^{2}+2^{2}+3^{2}+\ldots+n^{2}}{n !} \\
=\frac{\sum n^{2}}{n !}=\frac{n(n+1)(2 n+1)}{6 n !} \\
=\frac{1}{6}\left(\frac{2 n^{3}+3 n^{2}+n}{n !}\right) \\
=\frac{1}{6}\left(2 \cdot \frac{n^{3}}{n !}+\frac{3 n^{2}}{n !}+\frac{n}{n !}\right)
\end{array}
$$
$\therefore$ Sum of the series
$$
\begin{array}{l}
=\frac{1}{6}\left(2 \sum_{\mathrm{n}=1}^{\infty} \frac{\mathrm{n}^{3}}{\mathrm{n} !}+3 \sum_{\mathrm{n}=1}^{\infty} \frac{\mathrm{n}^{2}}{\mathrm{n} !}+\sum_{\mathrm{n}=1}^{\infty} \frac{\mathrm{n}}{\mathrm{n} !}\right) \\
=\frac{1}{6}(2 \times 5 \mathrm{e}+3 \times 2 \mathrm{e}+\mathrm{e}) \\
=\frac{1}{6}(10 \mathrm{e}+6 \mathrm{e}+\mathrm{e})=\frac{17}{6} \mathrm{e}
\end{array}
$$
Hence, the correct answer is option: 'B'.