Class 11MathematicsSequences and Seriesnth term of special series, Sum to n terms and Infinite number of terms
Medium$\sum\limits_{m = 1}^n {{m^2}}$ is equal to
- A$\frac{{m(m + 1)}}{2}$
- B$\frac{{m(m + 1)(2m + 1)}}{6}$
- C$\frac{{n(n + 1)(2n + 1)}}{6}$
- D$\frac{{n(n + 1)}}{2}$
Explore More
Similar Questions
The sum to infinite terms of the series $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$ is
DifficultAIEEE 2009
View Solution$2.\overline{357} = $
MediumIIT 1983
View SolutionThe sum of an infinite number of terms in a $G.P.$ is $20$ and the sum of their squares is $100$. The common ratio of the $G.P.$ is:
MediumAIEEE 2002
View SolutionIf the sum of $1 + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + \dots$ to $n$ terms is $S$,then $S$ is equal to
Easy
View SolutionIf $1^2 + 2^2 + 3^2 + \dots + 2009^2 = (2009)(335)(4019)$ and $(1)(2009) + 2(2008) + 3(2007) + \dots + 2009(1) = (2009)(335)(x)$,then $x$ is equal to:
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo