Q65 / 100 · BITSAT 2024 (Memory Based Paper 3) · Mathematics

The coefficient of $x^n$ in the expansion of $\frac{e^{7 x}+e^x}{e^{3 x}}$ is

  • A
    $\frac{4^{n-1}+(-2)^n}{n!}$
  • B
    $\frac{4^{n-1}+2^n}{n!}$
  • Correct answer
    C
    $\frac{4^n+(-2)^n}{n!}$
  • D
    $\frac{4^{n-1}+(-2)^{n-1}}{n!}$

Solution

Given, $\frac{e^{7 x}+e^x}{e^{3 x}} \Rightarrow e^{4 x}+e^{-2 x}$ Series of $e^a=1+\frac{a}{1!}+\frac{a^2}{2!}+\frac{a^3}{3!} \ldots$ $\begin{aligned} \Rightarrow e^{4 x}+ & e^{-2 x}=\left(1+\frac{4 x}{1!}+\frac{(4 x)^2}{2!}+\frac{(4 x)^3}{3!}+\ldots\right) \\ & +\left(1+\frac{(-2 x)}{1!}+\frac{(-2 x)^2}{2!}+\ldots \ldots\right) \end{aligned}$ Here, coefficient of $\begin{aligned} & x^2 \equiv \frac{4^2}{2!}+\frac{(-2)^2}{2!} \\ & x^3 \equiv \frac{4^3}{3!}+\frac{(-2)^3}{3!} \\ & \vdots \\ & x^n=\frac{4^n}{n!}+\frac{(-2)^n}{n!} \end{aligned}$

Hence, the correct answer is option: 'C'.

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