Question

The coefficient of \(x^n\) in the expansion of \[\frac{e^{7x} + e^x}{e^{3x}}\] is:

• A. \(\frac{4^{n-1} \cdot (-2)^n}{n!}\)
• B. \(\frac{4^n - 1 \cdot (2)^n}{n!}\)
• C. \(\frac{4^n + (-2)^n}{n!}\)
• D. \(\frac{4^n - 1 \cdot (-2)^{n-1}}{n!}\)

Hint:

When combining terms from exponential expansions, ensure consistent base and power adjustments to match given options.

Answer 1

The Correct Option is C

The expression is expanded using the Maclaurin series for \(e^x\). Given,
\[\frac{e^{7x} + e^x}{e^{3x}} \Rightarrow e^{4x} + e^{-2x}\] The Maclaurin series for \(e^a\) is \(1 + \frac{a}{1!} + \frac{a^2}{2!} + \frac{a^3}{3!} + \dots\).
Applying this to the expression:
\[\Rightarrow e^{4x} + e^{-2x} = \left(1 + \frac{4x}{1!} + \frac{(4x)^2}{2!} + \frac{(4x)^3}{3!} + \dots\right) + \left(1 + \frac{(-2x)}{1!} + \frac{(-2x)^2}{2!} + \dots\right)\] Comparing coefficients for powers of \(x\):
\[\begin{aligned}x^2 &\text{ coefficient} \equiv \frac{4^2}{2!} + \frac{(-2)^2}{2!} \\x^3 &\text{ coefficient} \equiv \frac{4^3}{3!} + \frac{(-2)^3}{3!} \\\vdots \\x^n &\text{ coefficient} = \frac{4^n}{n!} + \frac{(-2)^n}{n!}\end{aligned}\]This result corresponds to option (C), which correctly identifies the coefficient for \(x^n\).