1 + frac{1}{3!} + frac{1}{5!} + frac{1}{7!} + | vedclass

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

Question

(D) We know the expansion of $e^x$ is $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \infty$. Setting $x = 1$,we get $e = 1 +

Vedclass · Accepted Answer

$\frac{e - e^{-1}}{2}$

Vedclass · Answer

(D) We know the expansion of $e^x$ is $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \infty$. Setting $x = 1$,we get $e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots \infty$. Setting $x = -1$,we get $e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots \infty$. Subtracting the two series: $e - e^{-1} = (1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots) - (1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \dots) = 2(1 + \frac{1}{3!} + \frac{1}{5!} + \dots)$. Therefore,$1 + \frac{1}{3!} + \frac{1}{5!} + \dots = \frac{e - e^{-1}}{2}$.

$1 + \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + \dots \infty = $

Similar Questions

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

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

The sum of the series $\frac{4}{1!} + \frac{11}{2!} + \frac{22}{3!} + \frac{37}{4!} + \frac{56}{5!} + \dots$ is

In the expansion of $(1 + x + x^2)e^{-x}$,the coefficient of $x^2$ is

The value of $\frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \dots$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module