Class 11MathematicsSequences and Seriesnth term of special series, Sum to n terms and Infinite number of terms
MediumThe $n^{th}$ term of the series $\frac{1}{1} + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + \dots$ is:
- A$\frac{n + 1}{2}$
- B$\frac{n - 1}{2}$
- C$\frac{n^2 + 1}{2}$
- D$\frac{n^2 - 1}{2}$
Explore More
Similar Questions
The $n^{\text{th}}$ term of the series $1 + (3 + 5 + 7) + (9 + 11 + 13 + 15 + 17) + \ldots$ is:
DifficultAP EAMCET 2024
View Solution$1 + \frac{1^3 + 2^3}{1 + 2} + \frac{1^3 + 2^3 + 3^3}{1 + 2 + 3} + \dots + \frac{1^3 + 2^3 + 3^3 + \dots + 15^3}{1 + 2 + 3 + \dots + 15} - \frac{1}{2}(1 + 2 + 3 + \dots + 15)$ is equal to
DifficultJEE MAIN 2019
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:
If $a, x$ are real numbers and $|a| < 1, |x| < 1$,then $1 + (1+a)x + (1+a+a^2)x^2 + \dots \infty$ is equal to
DifficultWBJEE 2016
View SolutionIf the $n^{th}$ term of a series is $n(n + 1)$,then the sum of its $n$ terms is......
Easy
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