Binomial Theorem And Its Simple Applications Question 202
Question: The sum $ 1+\frac{1+a}{2!}+\frac{1+a+a^{2}}{3!}+…..\infty $ is equal to
Options:
A) $ e^{a} $
B) $ \frac{e^{a}-e}{a-1} $
C) $ (a-1)e^{a} $
D) $ (a+1)e^{a} $
Show Answer
Answer:
Correct Answer: B
Solution:
- [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}+…tonterms}{n!} $
$ =\frac{1(1-a^{n})}{(1-a)(n!)}=\frac{1}{1-a}( \frac{1-a^{n}}{n!} ) $
$ \therefore T_1+T_2+T_3+…..to\infty $
$ =\frac{1}{1-a}[ \frac{1-a}{1!}+\frac{1-a^{2}}{2!}+\frac{1-a^{3}}{3!}+….to\infty ] $
$ =\frac{1}{1-a}[ ( \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+…to\infty )-( \frac{a}{1!}+\frac{a^{2}}{2!}+\frac{a^{3}}{3!}+…to\infty ) ] $
$ =\frac{1}{1-a}[(e-1)-(e^{a}-1)]=\frac{e-e^{a}}{1-a}=\frac{e^{a}-e}{a-1} $
BETA
