Question

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

• A. $e$
• B. $2e$
• C. $\frac{3e}{2}$
• D. $\frac{4e}{5}$

Answer 1

The correct option is C

$\frac{3e}{2}$

Explanation for the correct answer:

The given sum is $S=1+\frac{\left(1+2\right)}{2!}+\frac{\left(1+2+3\right)}{3!}+\frac{\left(1+2+3+4\right)}{4!}+.......$

The $n^{th}$ term if this sum can be written as

$t_n=\frac{\left(1+2+3+...+n\right)}{n!}$

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

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

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

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

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

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

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

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

We know that

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

Substitute $x=1$

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

$\Rightarrow$ $S=1+\frac{e}{2}+e-1$

$\Rightarrow$ $S=\frac{3e}{2}$

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

Hence, option (C) is the correct answer.