A) \[e\]
B) \[{{e}^{-1}}\]
C) 1
D) 0
Correct Answer: D
Solution :
Given,\[\frac{1}{{{e}^{3x}}}({{e}^{x}}+{{e}^{5x}})={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...\] \[\Rightarrow \] \[({{e}^{-2x}}+{{e}^{2x}})={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...\] \[\Rightarrow \] \[2\left[ 1+\frac{{{(2x)}^{2}}}{2!}+\frac{{{(2x)}^{4}}}{4!}+... \right]\] \[={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...\] \[\Rightarrow \] \[{{a}_{1}}={{a}_{3}}={{a}_{5}}=...=0\] \[\therefore \] \[2{{a}_{1}}+{{2}^{3}}{{a}_{3}}+{{2}^{5}}{{a}_{5}}+...=0\]
You need to login to perform this action.
You will be redirected in
3 sec
