Question:medium
\( \frac{1}{e^{3x}}(e^{x} + e^{5x}) = a_{0} + a_{1}x + a_{2}x^{2} + \cdots \Rightarrow 2a_{1} + 2^{3}a_{3} + 2^{5}a_{5} + \cdots \) is equal to
\( \frac{1}{e^{3x}}(e^{x} + e^{5x}) = a_{0} + a_{1}x + a_{2}x^{2} + \cdots \Rightarrow 2a_{1} + 2^{3}a_{3} + 2^{5}a_{5} + \cdots \) is equal to
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An even function $f(x) = f(-x)$ has no odd powers in its Taylor expansion.
Updated On: Apr 10, 2026
- e
- $e^{-1}$
- 1
- 0
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The Correct Option is D
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