1 + frac{1 + x}{2!} + frac{1 + x + x^2}{3!} + | 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 $T_n = \frac{1 + x + x^2 + \dots + x^{n-1}}{n!}$.Using the sum of a geometric progression,$T_n = \frac{1

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The $n$-th term of the series is given by $T_n = \frac{1 + x + x^2 + \dots + x^{n-1}}{n!}$.Using the sum of a geometric progression,$T_n = \frac{1 - x^n}{1 - x} \cdot \frac{1}{n!} = \frac{x^n - 1}{(x - 1)n!} = \frac{1}{x - 1} \left( \frac{x^n}{n!} - \frac{1}{n!} \right)$.Summing from $n = 1$ to $\infty$,we get $\sum_{n=1}^{\infty} T_n = \frac{1}{x - 1} \left( \sum_{n=1}^{\infty} \frac{x^n}{n!} - \sum_{n=1}^{\infty} \frac{1}{n!} \right)$.Since $\sum_{n=1}^{\infty} \frac{x^n}{n!} = e^x - 1$ and $\sum_{n=1}^{\infty} \frac{1}{n!} = e - 1$,$\sum_{n=1}^{\infty} T_n = \frac{1}{x - 1} ((e^x - 1) - (e - 1)) = \frac{e^x - e}{x - 1}$.

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

Similar Questions

The sum of the series $\frac{1}{2 !} + \frac{1+2}{3 !} + \frac{1+2+3}{4 !} + \ldots$ is equal to :

The value of $1 - \log 2 + \frac{(\log 2)^2}{2!} - \frac{(\log 2)^3}{3!} + \dots$ is

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

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:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module