Question

$\frac{3}{1!}+\frac{5}{2!}+\frac{9}{3!}+\frac{15}{4!}+\frac{23}{5!}+.....\infty is$

• A. $4e-1$
• B. $4e-3$
• C. $3e+2$
• D. $3e+4$

Answer 1

The correct option is B

$4e-3$

Explanation for the correct option:

Step 1. Find the $nth$ term of given series:

Here, the numerator series are in AP

$\Rightarrow$$T_n=a{n}^2+bn+c$

Step 2. put $n=1,2,3$in $T_n$

$\Rightarrow$ $a+b+c=3$ …….(1)

$\Rightarrow$ $T_1=a+b+c$

$\Rightarrow$$4a+2b+c=5$ …….(2)

$\Rightarrow$ $T_2=4a+2b+c$

$\Rightarrow$$9a+3b+c=9$ ……..(3)

$\Rightarrow$ $T_3=9a+3b+c$

Step 3. solve equation (1), (2) and (3)

$\Rightarrow$ $T_n=n^2-n+3$

Step 4. Find the sum of the series:

$\Rightarrow \sum _{}\left(\frac{n^2-n+3}{n!}\right)$

$\Rightarrow \sum _{}\left(\frac{n^2}{n!}\right)–\sum _{}\left(\frac{n}{n!}\right)+\underset{}{3\sum }\left(\frac{1}{n!}\right)$

$\Rightarrow 2e-e+3(e-1)$; $[\because \frac{1}{n!}=(e-1),\frac{n}{n!}=e,\frac{n^2}{n!}=2e]$

$\Rightarrow 2e-e+3e-3$

$\Rightarrow \mathbf{4}\mathit{e}\mathbf{-}\mathbf{3}$

Hence, Option ‘B’ is Correct.