Sequence And Series Question 204
Question: The coefficient of $ x^{n} $ in the expansion of $ \frac{e^{7x}+e^{x}}{e^{3x}} $ is
[MP PET 1999]
Options:
A) $ \frac{{4^{n-1}}+{{(-2)}^{n}}}{n,!} $
B) $ \frac{{4^{n-1}}+2^{n}}{n,!} $
C) $ \frac{{4^{n-1}}+{{(-2)}^{n-1}}}{n,!} $
D) $ \frac{4^{n}+{{(-2)}^{n}}}{n,!} $
Show Answer
Answer:
Correct Answer: D
Solution:
We have $ \frac{e^{7x}+e^{x}}{e^{3x}}=e^{4x}+{e^{-2x}}=\sum\limits_{n=0}^{\infty }{\frac{{{(4x)}^{n}}}{n!}+\sum\limits_{n=0}^{\infty }{\frac{{{(-2x)}^{n}}}{n!}}} $
$ \therefore $ Coefficient of $ x^{n} $ in $ ( \frac{e^{7x}+e^{x}}{e^{3x}} )=\frac{4^{n}+{{(-2)}^{n}}}{n!} $ .
BETA
