Binomial Theorem And Its Simple Applications Question 131
Question: The co-efficient of $ x^{n} $ in the expansion of $ \frac{e^{7x}+e^{x}}{e^{3x}} $ is
Options:
A) $ \frac{{4^{n-1}}+{{(-2)}^{n}}}{n!} $
B) $ \frac{{4^{n-1}}+2^{n}}{n!} $
C) $ \frac{4^{n}+{{(-2)}^{n}}}{n!} $
D) $ \frac{{4^{n-1}}+{{(-2)}^{n-1}}}{n!} $
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Answer:
Correct Answer: C
Solution:
- [c] $ \frac{e^{7x}+e^{x}}{e^{3x}}=e^{4x}+{e^{-2x}} $
$ =[ 1+4x+\frac{{{(4x)}^{2}}}{2!}+… ]+[ 1+(-2x)+\frac{{{(-2x)}^{2}}}{2!}+… ] $
$ \therefore ,coeff.ofx^{n}=\frac{4^{n}}{n!}+\frac{{{(-2)}^{n}}}{n!} $
BETA
