Mathematics - Binomial Theorem Question with Solution | TestHub

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