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} $