Question

The sum of the following series $9+\frac{16}{2!}+\frac{27}{3!}+\frac{42}{4!}+...\infty$ is ae - b .Find $a+b$

Answer 1

Given, $y=9+\frac{16}{2!}+\frac{27}{3!}+\frac{42}{4!}+......\infty$

$\Rightarrow y=2+1+6+\frac{2\left(4\right)+2+6}{2!}+\frac{2\left(9\right)+3+6}{3!}+\frac{2\left(16\right)+4+6}{4!}+.......$

$\Rightarrow y=\sum _{n=1}^{\infty }{T}_n$ $\text{where,}{T}_n=\frac{2n^2+n+6}{n!}$

$\Rightarrow y=\sum _{n=1}^{\infty }[\frac{2n^2}{n!}+\frac{n}{n!}+\frac{6}{n!}]$

$\Rightarrow y=\sum _{n=1}^{\infty }[\frac{2n}{(n-1)!}+\frac{1}{(n-1)!}+\frac{6}{n!}]$

$\Rightarrow y=\frac{2}{(1-1)!}\sum _{n=2}^{\infty }\frac{2n-2+2}{(n-1)!}$$+\sum _{n=1}^{\infty }[\frac{1}{(n-1)!}+\frac{6}{n!}]$

$\Rightarrow y=2\sum _{n=2}^{\infty }\frac{2}{(n-2)!}$$+\sum _{n=2}^{\infty }\frac{2}{(n-1)!}$$+\sum _{n=1}^{\infty }[\frac{1}{(n-1)!}+\frac{6}{n!}]$

$\Rightarrow y=2+2[\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+....]+2[\frac{1}{1!}+\frac{1}{2!}+........]+[(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+....)+6(\frac{1}{1!}+\frac{1}{2!}+.....)]$

$\Rightarrow y=2+2\left[e\right]+2[e-1]+[e+6(e-1\left)\right]$

$\Rightarrow y=2+2e+2e-2+7e-6$

$\Rightarrow y=11e-6$

$\therefore a=11\text{and}b=6$

$\therefore a+b=17$