frac{2}{1!} + frac{4}{3!} + frac{6}{5!} + fra | vedclass

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

Question

(B) The given series is $S = \frac{2}{1!} + \frac{4}{3!} + \frac{6}{5!} + \frac{8}{7!} + \dots \infty$. We can write the $n$-th term as $\frac{2n}{(2n

Vedclass · Accepted Answer

$e$

Vedclass · Answer

(B) The given series is $S = \frac{2}{1!} + \frac{4}{3!} + \frac{6}{5!} + \frac{8}{7!} + \dots \infty$. We can write the $n$-th term as $\frac{2n}{(2n-1)!}$. $S = \sum_{n=1}^{\infty} \frac{2n}{(2n-1)!} = \sum_{n=1}^{\infty} \frac{(2n-1) + 1}{(2n-1)!} = \sum_{n=1}^{\infty} \left( \frac{2n-1}{(2n-1)!} + \frac{1}{(2n-1)!} \right)$. $S = \sum_{n=1}^{\infty} \frac{1}{(2n-2)!} + \sum_{n=1}^{\infty} \frac{1}{(2n-1)!}$. Expanding the terms: $S = \left( \frac{1}{0!} + \frac{1}{2!} + \frac{1}{4!} + \dots \right) + \left( \frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \dots \right)$. Recall that $e = \sum_{n=0}^{\infty} \frac{1}{n!}$ and $e^{-1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}$. Thus,$\frac{e + e^{-1}}{2} = \frac{1}{0!} + \frac{1}{2!} + \frac{1}{4!} + \dots$ and $\frac{e - e^{-1}}{2} = \frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \dots$. $S = \frac{e + e^{-1}}{2} + \frac{e - e^{-1}}{2} = \frac{2e}{2} = e$.

$\frac{2}{1!} + \frac{4}{3!} + \frac{6}{5!} + \frac{8}{7!} + \dots \infty = $

Similar Questions

In the expansion of $\frac{1 - 2x + 3x^2}{e^x}$,the coefficient of $x^5$ will be

$1 + \frac{3}{1!} + \frac{5}{2!} + \frac{7}{3!} + ....\infty = $

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

The sum of the series $\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$ 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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module