Q50 / 160 · TS EAMCET 2009 · Mathematics

$\frac{1}{e^{3 x}}\left(e^x+e^{5 x}\right)=a_0+a_1 x+a_2 x^2+\ldots$ $\Rightarrow \quad 2 a_1+2^3 a_3+2^5 a_5+\ldots$ is equal to

  • A
    $e$
  • B
    $e^{-1}$
  • C
    1
  • Correct answer
    D
    0

Solution

$\begin{aligned} & \text { Given, } \frac{1}{e^{3 x}}\left(e^x+e^{5 x}\right)=a_0+a_1 x+a_2 x^2+\ldots \\ & \Rightarrow\left(e^{-2 x}+e^{2 x}\right)=a_0+a_1 x+a_2 x^2+\ldots \\ & \Rightarrow \quad 2\left[1+\frac{(2 x)^2}{2 !}+\frac{(2 x)^4}{4 !}+\ldots\right] \\ & =a_0+a_1 x+a_2 x^2+\ldots \\ & \Rightarrow \quad a_1=a_3=a_5=\ldots=0 \\ & \therefore \quad 2 a_1+2^3 a_3+2^5 a_5+\ldots=0 \\ & \end{aligned}$

Hence, the correct answer is option: 'D'.

Relevant Questions