Class 11MathematicsSequences and Seriesnth term of special series, Sum to n terms and Infinite number of terms
MediumConsider a sequence whose sum of first $n$ terms is given by $S_n = 4n^2 + 6n$,where $n \in N$. Then,the $15^{th}$ term $(T_{15})$ of this sequence is:
- A$118$
- B$120$
- C$122$
- D$86$
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Similar Questions
If $\sum_{r=1}^{10} r! (r^3 + 6r^2 + 2r + 5) = \alpha(11!)$,then the value of $\alpha$ is equal to ...... .
DifficultJEE MAIN 2021
View SolutionFind the sum of the following series up to $n$ terms:
$\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots$
$\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots$
Difficult
View SolutionWhat is the sum of the series $1 + (1 + x) + (1 + x + x^2) + (1 + x + x^2 + x^3) + \dots$ up to $n$ terms?
Medium
View SolutionThe sum of the $1^{st} n$ 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:
DifficultKCET 2016
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:
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