In the expansion of frac{e^{4x} - 1}{e^{2x}}, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We are given the expression $\frac{e^{4x} - 1}{e^{2x}}$.Dividing the terms,we get $\frac{e^{4x}}{e^{2x}} - \frac{1}{e^{2x}} = e^{2x} - e^{-2x}$.Us

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) We are given the expression $\frac{e^{4x} - 1}{e^{2x}}$.Dividing the terms,we get $\frac{e^{4x}}{e^{2x}} - \frac{1}{e^{2x}} = e^{2x} - e^{-2x}$.Using the exponential series expansion $e^t = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \dots$,we have:$e^{2x} = 1 + (2x) + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots = 1 + 2x + 2x^2 + \frac{4x^3}{3} + \dots$$e^{-2x} = 1 + (-2x) + \frac{(-2x)^2}{2!} + \frac{(-2x)^3}{3!} + \dots = 1 - 2x + 2x^2 - \frac{4x^3}{3} + \dots$Subtracting these series: $(e^{2x} - e^{-2x}) = (1 - 1) + (2x - (-2x)) + (2x^2 - 2x^2) + \dots = 4x + 0x^2 + \frac{8x^3}{3} + \dots$The coefficient of $x^2$ is $0$.

In the expansion of $\frac{e^{4x} - 1}{e^{2x}}$,the coefficient of $x^2$ is

Similar Questions

$\sum_{n=1}^{\infty} \frac{2n^2+n+1}{n!}$ is equal to

In the expansion of $\frac{a + bx + cx^2}{e^x}$,the coefficient of $x^n$ is

The sum of the series $C = 1 + \frac{\cos x}{1!} + \frac{\cos 2x}{2!} + \frac{\cos 3x}{3!} + \dots$ and $S = \frac{\sin x}{1!} + \frac{\sin 2x}{2!} + \frac{\sin 3x}{3!} + \dots$ is equal to

$1 + \frac{2^4}{2!} + \frac{3^4}{3!} + \frac{4^4}{4!} + \dots \infty = $

The expression $\begin{aligned} & 1+x \log _e a+\frac{x^2}{2 !}\left(\log _e a\right)^2+\frac{x^3}{3 !}\left(\log _e a\right)^3+\ldots \end{aligned}$ for $a>0, x \in R$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module