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
If the $n^{th}$ term of a sequence is $T_n = 2n - 1$,then the sum of $n$ terms $S_n = \dots$
Easy
View SolutionThe numbers $a_n$ are defined by $a_0=1$ and $a_{n+1}=3n^2+n+a_n$ for $n \geq 0$. Then $a_n$ is equal to:
For any integer $n \geq 1$,the sum $\sum_{k=1}^n k(k+2)$ is equal to
DifficultAP EAMCET 2008
View SolutionIf $f(1)=3$, and $f(n+1)-f(n)=3(4^n-1)$, then for all $n \in N$, $f(n)=$
What is the sum of $n$ terms of the series $1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots$?
Medium
View SolutionVedclass 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