Question

The sum of the series $\frac{12}{2!}+\frac{28}{3!}+\frac{50}{4!}+\frac{78}{5!}+\dots \dots \dots \dots$ is

• A. e
• B. 3 e
• C. 4 e
• D. 5 e

Answer 1

Answer: (D)

$t_n=12+\frac{n-1}{2}[2\times 16+(n-2)\times 6]=3n^2+7n+2$
So, the general term of the given series
$T_n=\frac{3n^2+7n+2}{(n+1)!}=\frac{3\left(n^2+2n+1\right)+n-1}{(n+1)!}$
$=\frac{3(n+1)}{n!}+\frac{n+1-2}{(n+1)!}=\frac{3}{(n-1)!}+\frac{3}{n!}+\frac{1}{n!}-\frac{2}{(n+1)!}$
$=\frac{3}{(n-1)!}+\frac{4}{n!}-\frac{2}{(n+1)!}$
$\therefore$ Sum $=\sum _{n=1}^{\infty }{T}_n=3e+4(e-1)-2(e-2)=5e$