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

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

Let $\sum_{n=0}^{\infty} \frac{n^3((2n)!) + (2n-1)(n!)}{(n!)((2n)!)} = ae + \frac{b}{e} + c$,where $a, b, c \in \mathbb{Z}$ and $e = \sum_{n=0}^{\infty} \frac{1}{n!}$. Then $a^2 - b + c$ is equal to $................$.

If $y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$,then $x = $

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

If $a = \sum\limits_{n = 0}^\infty {\frac{{{x^{3n}}}}{{(3n)!}}} ,\,b = \sum\limits_{n = 1}^\infty {\frac{{{x^{3n - 2}}}}{{(3n - 2)!}}} $ and $c = \sum\limits_{n = 1}^\infty {\frac{{{x^{3n - 1}}}}{{(3n - 1)!}}} $ then the value of ${a^3} + {b^3} + {c^3} - 3abc = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module