Question

$\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+....+\infty$

• A. $12e$
• B. $12e-4$
• C. $12e-3$
• D. $12e-5$

Answer 1

The correct option is D

$12e-5$

Explanation for correct options:

Step1: Finding general term of the given series.

Given, $\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+....+\infty$

We use the formula:

$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\infty$

Put $x=1$

$e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+.....+\infty$

Now consider the numerators of the series$9,19,35,57\dots$

Let an AP with $a=10andd=6$

$\begin{array}{rcl}{t}_n& =& 9+\left(\frac{(n-1)}{2}\right)(20+\frac{(n-1)}{6})\\ t_n& =& 9+3n^2+4n-3n-4\\ t_n& =& 3n^2+n+5\end{array}$

Therefore ,

$T_n=\frac{t_n}{n!}=\frac{3n^2+n+5}{n!}$

Step 2: Finding sum of the $T_n$

General Term of the given summation is $T_{n=}\frac{3n^2+n+5}{n!}$

Now,

$\sum _{}{T}_n=\sum _{}^{}\frac{3n^2+n+5}{n!}$

$=\sum _{}^{}(\frac{3}{(n-2)!}+\frac{3}{(n-1)!}+\frac{1}{(n-1)!}+\frac{5}{n!})$

$=\sum _{n=2}^{\infty }\frac{3}{(n-2)!}+\sum _{n=1}^{\infty }\frac{4}{(n-1)!}+\sum _{n=1}^{\infty }\frac{5}{n!}$

$=3(1+\frac{1}{1}+\frac{1}{2}!+...)+4(1+\frac{1}{1}+\frac{1}{2}!+...)+5(1+\frac{1}{1}+\frac{1}{2}!+...)$

$$\begin{gathered} =3e+4e+5(e-1) \\ =12e–5 \end{gathered}$$

Hence, Option(D) is correct.