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) \[\frac{{{4}^{n}}+{{(-2)}^{n}}}{n\,!}\]
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 \[\left( \frac{{{e}^{7x}}+{{e}^{x}}}{{{e}^{3x}}} \right)=\frac{{{4}^{n}}+{{(-2)}^{n}}}{n!}\].
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