Question

If the sum of the series $\frac{3}{1!}+\frac{5}{2!}+\frac{7}{3!}+\frac{9}{4!}+...\infty =Ae+B$
Find the value of $A+B$

• A. $1$
• B. $7$
• C. $0$
• D. $Noneofthese$

Answer 1

The correct option is A $1$

Its general term is $\frac{2n+1}{n!}$

So we have to calculate ${\sum }_{n=0}^{n=\infty }\frac{2n+1}{n!}=2{\sum }_{k=0}^{k=\infty }\frac{1}{k!}+{\sum }_{n=0}^{n=\infty }\frac{1}{n!}-2=3e-2$ (using taylor's expansion for $e^x$)

So A+B=1