Question

$\frac{1}{e^{3 x}}\left(e^{x}+e^{5 x}\right)=a_{0}+a_{1} x +a_{2} x^{2}+\ldots$ $\Rightarrow 2 a_{1}+2^{3} a_{3}+2^{5} a_{5}+\ldots$ is equal to

• A. $e$
• B. $e^{-1}$
• C. 1
• D. 0

Answer 1

Answer: (D)

0

Answer 2

The correct option is(D): 0.

Given, $\frac{1}{e^{3 x}}\left(e^{x}+e^{5 x}\right)=a_{0}+a_{1} x+ a_{2} x^{2}+\ldots$
$\Rightarrow \left(e^{-2 x}+e^{2 x}\right)=a_{0}+a_{1} x+ a_{2} x^{2}+\ldots$
$\Rightarrow 2\left[1+\frac{(2 x)^{2}}{2 !}+\frac{(2 x)^{4}}{4 !}+\ldots\right]$
$=a_{0}+a_{1} x+ a_{2} x^{2}+\ldots$
$\Rightarrow a_{1}=a_{3}=a_{5}=\ldots=0$
$\therefore 2 a_{1}+2^{3} a_{3}+2^{5} a_{5}+\ldots=0$