A) \[\frac{{{4}^{n-1}}+{{(-2)}^{n}}}{n!}\]
B) \[\frac{{{4}^{n-1}}+{{2}^{n}}}{n!}\]
C) \[\frac{{{4}^{n}}+{{(-2)}^{n}}}{n!}\]
D) \[\frac{{{4}^{n-1}}+{{(-2)}^{n-1}}}{n!}\]
Correct Answer: C
Solution :
[c] \[\frac{{{e}^{7x}}+{{e}^{x}}}{{{e}^{3x}}}={{e}^{4x}}+{{e}^{-2x}}\] \[=\left[ 1+4x+\frac{{{(4x)}^{2}}}{2!}+... \right]+\left[ 1+(-2x)+\frac{{{(-2x)}^{2}}}{2!}+... \right]\] \[\therefore \,\,\,\,\,\,\,coeff.\text{ }of\,\,{{x}^{n}}=\frac{{{4}^{n}}}{n!}+\frac{{{(-2)}^{n}}}{n!}\]
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