Question

The sum of the infinite series $1+\frac{1+2}{2!}+\frac{1+2+2^2}{3!}+\frac{1+2+2^2+2^3}{4!}+...$ is $e^y-e^x$ Find $x+y^2$

Answer 1

$S=1+\frac{1+2}{2!}+\frac{1+2+2^2}{3!}+\frac{1+2+2^2+2^3}{4!}+...$
Now, $t_n=\frac{2^n-1}{n!}$
$S=\sum _{n=0}^{\infty }{t}_n=\sum _{n=0}^{\infty }\frac{2^n}{n!}-\sum _{n=0}^{\infty }\frac{1}{n!}=e^2-e$
Therefore, $x+y^2=5$
Ans: 5