Question

The sum of the series $\frac{3}{1!}+\frac{5}{2!}+\frac{9}{3!}+\frac{15}{4!}+\frac{23}{5!}+...\infty$ is $(a\ast e)-b$.Find $a\times b$

Answer 1

$I=\frac{3}{1!}+\frac{5}{2!}+\frac{9}{3!}+\frac{15}{4!}+\frac{23}{5!}+...\infty =\sum _{n=0}^{\infty }\frac{2n+1}{n!}=2\sum _{n=1}^{\infty }\frac{1}{(n-1)!}+\sum _{n=0}^{\infty }\frac{1}{n!}=2(e-1)+e=3e-2$
Ans: 6