Question

Let $f_n\left(x\right)=\frac{n+1}{(n-1)!}{x}^n$. The value of ${\int }_0^1\left(\sum _{n=1}^{\infty }{f}_n\right(x\left)\right)dx$ is

• A. $e$
• B. $0$
• C. $2e$
• D. none of these

Answer 1

Answer: (A)

$e$

To find the value of ${\int }_0^1\left(\sum _{n=1}^{\infty }{f}_n\right(x\left)\right)dx$

Now, ${\sum }_{n=1}^{\infty }{f}_n\left(x\right)=2x+3x^2+\frac{4x^3}{2!}+\frac{5x^4}{3!}+...\infty$
The series can be written as
$\frac{d\left(x^2{e}^x\right)}{dx}$
Hence,
${\int }_0^1\frac{d\left(x^2{e}^x\right)}{dx}.dx$
$=[x^2{e}^x{]}_0^1$
$=e$