Question:
The coefficient of \(x^n\) in the expansion of \[\frac{e^{7x} + e^x}{e^{3x}}\] is:
The coefficient of \(x^n\) in the expansion of \[\frac{e^{7x} + e^x}{e^{3x}}\] is:
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When combining terms from exponential expansions, ensure consistent base and power adjustments to match given options.
Updated On: Jan 13, 2026
- \(\frac{4^{n-1} \cdot (-2)^n}{n!}\)
- \(\frac{4^n - 1 \cdot (2)^n}{n!}\)
- \(\frac{4^n + (-2)^n}{n!}\)
- \(\frac{4^n - 1 \cdot (-2)^{n-1}}{n!}\)
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The Correct Option is C
Solution and Explanation
The given expression can be expanded using the Maclaurin series for \(e^x\):
Given,
\[\frac{e^{7x} + e^x}{e^{3x}} \Rightarrow e^{4x} + e^{-2x}\] {Series of } \(e^a = 1 + \frac{a}{1!} + \frac{a^2}{2!} + \frac{a^3}{3!} + \dots\)
\[\Rightarrow e^{4x} + e^{-2x} = \left(1 + \frac{4x}{1!} + \frac{(4x)^2}{2!} + \frac{(4x)^3}{3!} + \dots\right) + \left(1 + \frac{(-2x)}{1!} + \frac{(-2x)^2}{2!} + \dots\right)\] \[ \begin{aligned} x^2 &\equiv \frac{4^2}{2!} + \frac{(-2)^2}{2!} \\ x^3 &\equiv \frac{4^3}{3!} + \frac{(-2)^3}{3!} \\ \vdots \\ x^n &= \frac{4^n}{n!} + \frac{(-2)^n}{n!} \end{aligned} \] This matches option (C), providing the correct coefficient for \(x^n\).
\[\frac{e^{7x} + e^x}{e^{3x}} \Rightarrow e^{4x} + e^{-2x}\] {Series of } \(e^a = 1 + \frac{a}{1!} + \frac{a^2}{2!} + \frac{a^3}{3!} + \dots\)
\[\Rightarrow e^{4x} + e^{-2x} = \left(1 + \frac{4x}{1!} + \frac{(4x)^2}{2!} + \frac{(4x)^3}{3!} + \dots\right) + \left(1 + \frac{(-2x)}{1!} + \frac{(-2x)^2}{2!} + \dots\right)\] \[ \begin{aligned} x^2 &\equiv \frac{4^2}{2!} + \frac{(-2)^2}{2!} \\ x^3 &\equiv \frac{4^3}{3!} + \frac{(-2)^3}{3!} \\ \vdots \\ x^n &= \frac{4^n}{n!} + \frac{(-2)^n}{n!} \end{aligned} \] This matches option (C), providing the correct coefficient for \(x^n\).
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