Question

$1+\frac{1+a}{2!}+\frac{1+a+a^2}{3!}+...\infty =\frac{e^a-e}{a-b}$ Find b

Answer 1

In the L.H.S.first term contains one term, second term contains two terms, third term contains three terms,..., nth term contains n terms.
$n$th Term
$T_n=\frac{1+a+a^2+..+a^{n-1}}{n!}$

$=\frac{1.(1-a^n)}{(1-a)n!}$= $\frac{1}{(1-a)}(\frac{1}{n!}-\frac{a^n}{n!})$
Hence sum of the series
$S=\sum _{n=1}^{\infty }{T}_n$
$=\sum _{n=1}^{\infty }\frac{1}{(1-a)}(\frac{1}{n!}-\frac{a^n}{n!})$
$=\frac{1}{(1-a)}(\sum _{n=1}^{\infty }\frac{1}{n!}-\sum _{n=1}^{\infty }\frac{a^n}{n!})$

$=\frac{1}{(1-a)}((\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...\infty )-(\frac{a}{1!}+\frac{a^2}{2!}+\frac{a^3}{3!}+...\infty ))$
$=\frac{1}{(1-a)}\left(\right(e-1)-(e^a-1\left)\right)=\frac{e^a-e}{(a-1)}$