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

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

If ${T_n} = \frac{{{3^n}}}{{2(n!)}} - \frac{1}{{2(n!)}}$,then ${S_\infty } = $

In the expansion of $(e^x - 1)(e^{-x} + 1)$,the coefficient of $x^3$ is

$b = 1 + \frac{{}^1 C_0 + {}^1 C_1}{1!} + \frac{{}^2 C_0 + {}^2 C_1 + {}^2 C_2}{2!} + \frac{{}^3 C_0 + {}^3 C_1 + {}^3 C_2 + {}^3 C_3}{3!} + \ldots$
Let $a = 1 + \frac{{}^2 C_2}{3!} + \frac{{}^3 C_2}{4!} + \frac{{}^4 C_2}{5!} + \ldots$. Then $\frac{2b}{a^2}$ is equal to:

The coefficient of $x^{10}$ in the expansion of $(2+3x)e^{-x}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module