TS EAMCET · Maths · Binomial Theorem
\(\frac{1}{e^{3 x}}\left(e^x+e^{5 x}\right)=a_0+a_1 x+a_2 x^2+\ldots\) \(\Rightarrow \quad 2 a_1+2^3 a_3+2^5 a_5+\ldots\) is equal to
- A \(e\)
- B \(e^{-1}\)
- C 1
- D 0
Answer & Solution
Correct Answer
(D) 0
Step-by-step Solution
Detailed explanation
\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…
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