Question

If \[ \frac{e^x+e^{5x}}{e^{3x}}=a_0+a_1x+a_2x^2+a_3x^3+\cdots, \] then the value of \(2a_1+2^3a_3+2^5a_5+\cdots\) is

• A. \(e^2+e^{-2}\)
• B. \(e^4-e^{-4}\)
• C. \(e^4+e^{-4}\)
• D. \(0\)

Hint:

Even functions contain only even powers.

Answer 1

The Correct Option is A

Step 1: \[ \frac{e^x+e^{5x}}{e^{3x}}=e^{-2x}+e^{2x} \]
Step 2: Maclaurin expansion: \[ e^{2x}+e^{-2x}=2\left(1+\frac{(2x)^2}{2!}+\frac{(2x)^4}{4!}+\cdots\right) \]
Step 3: Odd power coefficients vanish.
Step 4: \[ 2a_1+2^3a_3+2^5a_5+\cdots=e^2+e^{-2} \]