Question

$1+\frac{1+2}{2!}+\frac{1+2+2^2}{3!}+\frac{1+2+2^2+2^3}{4!}\dots .=$

• A. $e^3$
• B. $e^2$
• C. $e^2-e$
• D. $e^3-e$

Answer 1

Answer: (C)

$e^2-e$

$1+\frac{1+2}{2!}+\frac{1+2+2^2}{3!}+\frac{1+2+2^2+2^3}{4!}\dots$

$=\sum _{n=1}^{\infty }\frac{1+2^1+2^2\cdots 2^{n-1}}{n!}$

$=\sum _{n=1}^{\infty }\frac{2^n-1}{n!}$ .... $[1+2^1+2^2+....$ is the geometric sum of $n$ terms with common ratio $2]$

$=\sum _{n=1}^{\infty }\frac{2^n}{n!}-\sum _{n=1}^{\infty }\frac{1}{n!}$

The first series is the expansion of $e^2$, whereas the second one is the series expansion of $e$.

and we know, $e^x=1+x+\frac{x^2}{2!}+\frac{x3}{3!}+...=\sum _{n=1}^{\infty }\frac{x^n}{n!}$

Hence we get

$1+\frac{1+2}{2!}+\frac{1+2+2^2}{3!}+\frac{1+2+2^2+2^3}{4!}\dots$$=e^2-e$