Question

The value of $\frac{2}{1!}+\frac{\left(2+4\right)}{2!}+\frac{\left(2+4+6\right)}{3!}+....$ is

• A. $e$
• B. $2e$
• C. $3e$
• D. none of these

Answer 1

The correct option is C

$3e$

Explanation for the correct answer:

Let $S=$$\frac{2}{1!}+\frac{\left(2+4\right)}{2!}+\frac{\left(2+4+6\right)}{3!}+....$

$=\frac{2}{1!}+\frac{6}{2!}+\frac{12}{3!}+.......$

The $n^{th}$ term of this summation can be written as

$t_n=\frac{n\left(n+1\right)}{n!}$

$\Rightarrow$ $t_n=\frac{\left(n+1\right)}{\left(n-1\right)!}$

$\Rightarrow$ $t_n=\frac{\left(n-1+2\right)}{\left(n-1\right)!}=\frac{n-1}{\left(n-1\right)!}+\frac{2}{\left(n-1\right)!}$

$\Rightarrow$ $t_n=\frac{1}{\left(n-2\right)!}+\frac{2}{\left(n-1\right)!}$

We cannot substitute $n=1$ as $\left(-1\right)!$ is not defined

$\therefore$ $S=\frac{2}{1!}+\sum _{i=2}^n{t}_n$

$\Rightarrow$ $S=2+\sum _{i=2}^n\frac{1}{(n-2)!}+2\sum _{i=2}^n\frac{1}{\left(n-1\right)!}$

$\Rightarrow$ $S=2+\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+....+2\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+....\right)$

We know that

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

Substitute $x=1$

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

$\Rightarrow$ $S=2+e+2\left(e-1\right)$

$\Rightarrow$ $S=3e$

Hence, the value of $\frac{2}{1!}+\frac{\left(2+4\right)}{2!}+\frac{\left(2+4+6\right)}{3!}+....$ is $3e$.

Hence, option $\left(C\right)$ is the correct answer.