Question

The sum $1+\frac{1+a}{2!}+\frac{1+a+a^2}{3!}+...\infty$ is equal to

• A. $e^a$
• B. $\frac{e^a-e}{a-1}$
• C. $(a-1)e^a$
• D. $(a+1)e^a$

Answer 1

Answer: (B)

The given series is
$1+\frac{1+a}{2!}+\frac{1+a+a^2}{3!}+\frac{1+a+a^2+a^3}{4!}+.....$
Here , $T_n=\frac{1+a+a^2+a^3+...\text{to}n\text{terms}}{n!}$
$=\frac{1\left(1-a^n\right)}{\left(1-a\right)\left(n!\right)}=\frac{1}{1-a}\left(\frac{1-a^n}{n!}\right)$
$\therefore T_1+T_2+T_3+....\text{to }\infty$
$=\frac{1}{1-a}\left[\frac{1-a}{1!}+\frac{1-a^2}{2!}+\frac{1-a^3}{3!}+...\text{to }\infty \right]$
$=\frac{1}{1-a}\left[\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...\text{to}\infty \right)-\left(\frac{a}{1!}+\frac{a^2}{2!}+\frac{a^3}{3!}+....\text{to}\infty \right)\right]$
$=\frac{1}{1-a}\left[\left(e-1\right)-\left(e^a-1\right)\right]$
$=\frac{e-e^a}{1-a}=\frac{e^a-e}{a-1}$