If $S$ is the sum of $n$ terms of the series $1 + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + \dots$,then $S = \dots$

  • A
    $\frac{n(n + 3)}{4}$
  • B
    $\frac{n(n + 2)}{4}$
  • C
    $\frac{n(n + 1)(n + 2)}{6}$
  • D
    $n^2$

Explore More

Similar Questions

The sum,$\sum_{n=1}^{7} \frac{n(n+1)(2n+1)}{4}$ is equal to

The $99^{th}$ term of the series $2 + 7 + 14 + 23 + 34 + \dots$ is

Difficult
View Solution

The value of the expression $1.(2 - \omega )(2 - {\omega ^2}) + 2.(3 - \omega )(3 - {\omega ^2}) + ....... + (n - 1).(n - \omega )(n - {\omega ^2}),$ where $\omega$ is an imaginary cube root of unity,is

If $1^2 + 2^2 + 3^2 + \dots + 2009^2 = (2009)(335)(4019)$ and $(1)(2009) + 2(2008) + 3(2007) + \dots + 2009(1) = (2009)(335)(x)$,then $x$ is equal to:

$\frac{1^3 + 2^3 + 3^3 + 4^3 + \dots + 12^3}{1^2 + 2^2 + 3^2 + 4^2 + \dots + 12^2} = $

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