Question

The sum of the series$\frac{1^2{.2}^2}{1!}+\frac{2^2{.3}^2}{2!}+\frac{3^2{.4}^2}{3!}+\dots \dots$ upto infinity is

• A. $27e$
• B. $24e$
• C. $28e$
• D. $e$

Answer 1

Answer: (A)

$27e$

Given, ${\sum }_{r=1}^{\infty }\frac{r^2{(r+1)}^2}{r!}$

$={\sum }_{r=1}^{\infty }\frac{r^4+2r^3+r^2}{r!}$

$={\sum }_{r=1}^{\infty }\frac{r^2}{r!}+\frac{2r^3}{r!}+\frac{r^4}{r!}$

${\sum }_{r=0}^{\infty }\frac{x^r}{r!}=e^x={\sum }_{r=1}^{\infty }\frac{x^{r-1}}{(r-1)!}$

${\sum }_{r=1}^{\infty }\frac{r^2}{r(r-1)!}={\sum }_{r=1}^{\infty }\frac{(r-1)}{(r-1)(r-2)!}+\frac{1}{(r-1)!}$

$={\sum }_{r=2}^{\infty }\frac{1}{(r-2)!}+{\sum }_{r=1}^{\infty }\frac{1}{(r-1)!}$

$=2e$

${\sum }_{r=1}^{\infty }\frac{2r^3}{r!}={\sum }_{r=1}^{\infty }\frac{2r^2}{(r-1)!}$

$={\sum }_{r=1}^{\infty }\frac{2(r^2-1)+2}{(r-1)!}$

$={\sum }_{r=2}^{\infty }\frac{2(r+1)}{(r-2)!}+2{\sum }_{r=1}^{\infty }\frac{1}{(r-1)!}$

$=2\left[{\sum }_{r=2}^{\infty }\frac{(r-2)}{(r-2)!}+\frac{3}{(r-2)!}\right]+2e$

$=2{\sum }_{r=3}^{\infty }\frac{1}{(r-3)!}+6{\sum }_{r=2}^{\infty }\frac{1}{(r-2)!}+2e$

$=2e+6e+2e=10e$

${\sum }_{r=1}^{\infty }\frac{r^4}{r!}={\sum }_{r=1}^{\infty }\frac{r^3}{(r-1)!}$

$={\sum }_{r=1}^{\infty }\frac{(r^3-1)}{(r-1)!}+{\sum }_{r=1}^{\infty }\frac{1}{(r-1)!}$

$={\sum }_{r=2}^{\infty }\frac{(r^2+r+1)}{(r-2)!}+e$

$={\sum }_{r=2}^{\infty }\frac{r^2-4}{(r-2)!}+\frac{(r-2)}{(r-2)!}+\frac{7}{(r-2)!}+e$

$={\sum }_{r=3}^{\infty }\frac{r+2}{(r-3)!}+{\sum }_{r=3}^{\infty }\frac{1}{(r-3)!}+7e+e$

$={\sum }_{r=3}^{\infty }(\frac{r-3}{(r-3)!}+\frac{5}{(r-3)!})+e+8e$

$={\sum }_{r=4}^{\infty }\frac{1}{(r-4)!}+5e+9e$

$=e+14e=15e$

${\sum }_{r=1}^{\infty }\frac{r^2{(r+1)}^2}{r!}=15e+10e+2e=27e$