Question:
\( \frac{1}{e^{3x}}(e^{x} + e^{5x}) = a_{0} + a_{1}x + a_{2}x^{2} + \cdots \Rightarrow 2a_{1} + 2^{3}a_{3} + 2^{5}a_{5} + \cdots \) 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
Solution and Explanation
Step 1: Simplifying Expression
$e^{-3x}(e^x + e^{5x}) = e^{-2x} + e^{2x}$.
Step 2: Symmetry Analysis
$e^{2x} + e^{-2x} = 2 \cosh(2x) = 2[1 + \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} + ...]$.
Step 3: Coefficients
In the expansion, only even powers of $x$ exist. This means all odd-indexed coefficients ($a_1, a_3, a_5, ...$) are zero.
Step 4: Result
Since $a_1 = a_3 = a_5 = 0$, the sum $2a_1 + 2^3a_3 + 2^5a_5 + ... = 0$.
Final Answer: (d)
$e^{-3x}(e^x + e^{5x}) = e^{-2x} + e^{2x}$.
Step 2: Symmetry Analysis
$e^{2x} + e^{-2x} = 2 \cosh(2x) = 2[1 + \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} + ...]$.
Step 3: Coefficients
In the expansion, only even powers of $x$ exist. This means all odd-indexed coefficients ($a_1, a_3, a_5, ...$) are zero.
Step 4: Result
Since $a_1 = a_3 = a_5 = 0$, the sum $2a_1 + 2^3a_3 + 2^5a_5 + ... = 0$.
Final Answer: (d)
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