Question

The sum of $\frac{1^2·2}{1!}+\frac{2^2·3}{2!}+\frac{3^2·4}{3!}+..............\infty$

• A. $5e$
• B. $3e$
• C. $7e$
• D. $2e$

Answer 1

The correct option is C

$7e$

Explanation for the correct option

Given: $\frac{1^2·2}{1!}+\frac{2^2·3}{2!}+\frac{3^2·4}{3!}+..............\infty$

$\frac{1^2·2}{1!}+\frac{2^2·3}{2!}+\frac{3^2·4}{3!}+..............\infty$

$n^{th}$ term$\left(T_n\right)$

$$\begin{gathered} \Rightarrow T_n=\frac{n^2(n+1)}{n!} \\ \Rightarrow T_n=\frac{n(n+1)}{(n-1)!} \\ \Rightarrow T_n=\frac{(n+2)+{\displaystyle \frac{2}{n-1}}}{(n-2)!} \\ \Rightarrow T_n=\frac{n+2}{(n-2)!}+\frac{2}{(n-1)(n-2)!} \\ \Rightarrow T_n=\frac{n+4-2}{(n-2)!}+\frac{2}{(n-1)(n-2)!} \\ \Rightarrow T_n=\frac{n-2}{(n-2)(n-3)!}+\frac{4}{(n-2)!}+\frac{2}{(n-1)!} \\ \Rightarrow T_n=\frac{1}{(n-3)!}+\frac{4}{(n-2)!}+\frac{2}{(n-1)!} \end{gathered}$$

put $n=3$ and formatting $T_3$ in the following form

$$\begin{gathered} \Rightarrow T_3=\frac{1}{(3-3)!}+\frac{4}{(3-2)!}+\frac{2}{(3-1)!} \\ \Rightarrow T_3=\frac{1}{0!}+\frac{4}{1!}+\frac{2}{2!} \end{gathered}$$

similarly,

$T_4=\frac{1}{1}+\frac{2}{3!}+\frac{4}{2!}$

$T_5=\frac{1}{2!}+\frac{2}{4!}+\frac{4}{3!}$

$$\begin{gathered} T_1=1^2·\frac{2}{1!}\Rightarrow \frac{2}{1!} \\ {T}_2=2^2·\frac{3}{2!}\Rightarrow 6\Rightarrow \frac{2}{0!}+\frac{4}{0!} \end{gathered}$$

So,

$$\begin{gathered} T_1+T_2+T_3+.........=\frac{2}{1!}+\frac{2}{0!}+\frac{4}{0!}+\frac{1}{0!}+\frac{4}{1!}+\frac{2}{2!}+................ \\ {T}_1+T_2+T_3+......=\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+.........\right)+2\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+.........\right)1+4\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+.........\right)---------\left(1\right) \end{gathered}$$

the expansion for $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+.........$

put $x=1$

$$\begin{gathered} e^1=1+\frac{1}{1!}+\frac{1^2}{2!}+\frac{1^3}{3!}+......... \\ e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+......... \end{gathered}$$

put the value in $\left(1\right)$

$$\begin{gathered} T_1+T_2+T_3+.........=e+2e+4e \\ {T}_1+T_2+T_3+.........=7e \end{gathered}$$

Hence, option C is correct.