The sum of the first $n$ terms of the series $\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+\ldots$ is

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

Explore More

Similar Questions

If the set of natural numbers is partitioned into subsets $S_1 = \{1\}, S_2 = \{2, 3\}, S_3 = \{4, 5, 6\}$ and so on,then the sum of the terms in $S_{50}$ is

Difficult
View Solution

If $0 < \theta, \phi < \frac{\pi}{2}$,$x = \sum_{n=0}^{\infty} \cos^{2n} \theta$,$y = \sum_{n=0}^{\infty} \sin^{2n} \phi$,and $z = \sum_{n=0}^{\infty} \cos^{2n} \theta \cdot \sin^{2n} \phi$,then:

Let $S$ be the infinite sum given by $S = \sum_{n=0}^{\infty} \frac{a_n}{10^{2n}}$,where $(a_n)_{n \geq 0}$ is a sequence defined by $a_0 = 1, a_1 = 1$ and $a_j = 20a_{j-1} - 108a_{j-2}$ for $j \geq 2$. If $S$ is expressed in the form $\frac{a}{b}$,where $a$ and $b$ are coprime positive integers,then $a$ equals:

The sum of an infinite geometric series is $3$. $A$ series,which is formed by squaring its terms,also has a sum of $3$. The first series is

The sum of infinite terms of the geometric progression $\frac{\sqrt{2} + 1}{\sqrt{2} - 1}, \frac{1}{2 - \sqrt{2}}, \frac{1}{2}, \dots$ is

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