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