1 + frac{1 + 2}{2!} + frac{1 + 2 + 3}{3!} + f | vedclass

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

Question

(D) The $n^{th}$ term of the series is given by $T_n = \frac{1 + 2 + 3 + \dots + n}{n!} = \frac{n(n + 1)}{2 \cdot n!} = \frac{n(n + 1)}{2 \cdot n(n -

Vedclass · Accepted Answer

$3e/2$

Vedclass · Answer

(D) The $n^{th}$ term of the series is given by $T_n = \frac{1 + 2 + 3 + \dots + n}{n!} = \frac{n(n + 1)}{2 \cdot n!} = \frac{n(n + 1)}{2 \cdot n(n - 1)!} = \frac{n + 1}{2(n - 1)!}$.We can rewrite the numerator as $(n - 1) + 2$,so $T_n = \frac{1}{2} \left[ \frac{n - 1}{(n - 1)!} + \frac{2}{(n - 1)!} \right] = \frac{1}{2} \left[ \frac{1}{(n - 2)!} + \frac{2}{(n - 1)!} \right]$.The sum of the series is $S = \sum_{n=1}^{\infty} T_n = \frac{1}{2} \left[ \sum_{n=1}^{\infty} \frac{1}{(n - 2)!} + \sum_{n=1}^{\infty} \frac{2}{(n - 1)!} \right]$.Note that $\frac{1}{(n-2)!} = 0$ for $n=1$,so the first sum starts from $n=2$,which is $\sum_{k=0}^{\infty} \frac{1}{k!} = e$.The second sum is $2 \sum_{n=1}^{\infty} \frac{1}{(n-1)!} = 2 \sum_{k=0}^{\infty} \frac{1}{k!} = 2e$.Thus,$S = \frac{1}{2} [e + 2e] = \frac{3e}{2}$.

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

Similar Questions

The sum of the infinite series $1 + 2 + \frac{1}{2!} + \frac{2}{3!} + \frac{1}{4!} + \frac{2}{5!} + \dots$ is

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

The coefficient of $x^k$ in the expansion of $\frac{1-2x-x^2}{e^{-x}}$ is

In the expansion of $(e^x - 1)^2$,the coefficient of $x^4$ will be

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module