Find the sum of the series: $1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \dots + n \cdot n!$

  • A
    $(n + 1)! - 1$
  • B
    $(n + 1)! + 1$
  • C
    $n! - 1$
  • D
    $n! + 1$

Explore More

Similar Questions

The sum of the series $\frac{1}{{1 + {1^2} + {1^4}}} + \frac{2}{{1 + {2^2} + {2^4}}} + \frac{3}{{1 + {3^2} + {3^4}}} + \dots$ to $n$ terms is

Difficult
View Solution

The value of $\sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\frac{3}{{{r^2} - r + 9}}} \right)} $ is-

The value of $\mathop {\lim }\limits_{n \to \infty } \left( \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2n - 1)(2n + 1)} \right)$ is equal to

The value of $\sum_{k=1}^{13} \frac{1}{\sin \left(\frac{\pi}{4}+\frac{(k-1) \pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)}$ is equal to

If $S$ is the sum of the first $10$ terms of the series $\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right)+\tan ^{-1}\left(\frac{1}{21}\right)+\ldots$ then $\tan ( S )$ is equal to

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo