Class 11MathematicsSequences and Seriesnth term of special series, Sum to n terms and Infinite number of terms
MediumIf the sum of the first $n$ terms of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \dots$ is $\frac{n(n+1)^2}{2}$ when $n$ is even,what is the sum when $n$ is odd?
- A$\frac{n^2(n+1)}{2}$
- B$\frac{n(n+1)(2n+1)}{6}$
- C$\frac{n(n+1)^2}{2}$
- D$\frac{n^2(n+1)^2}{2}$
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Let $a_{n}$ be the $n^{\text{th}}$ term of the series $5+8+14+23+35+50+\ldots$ and $S_{n}=\sum_{k=1}^{n} a_{k}$. Then $S_{30}-a_{40}$ is equal to
DifficultJEE MAIN 2023
View SolutionThe sum to infinity of the following series $2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \dots$ is:
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View SolutionFind the sum of the following series up to $n$ terms:
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Difficult
View Solution$\sum\limits_{i=1}^n \sum\limits_{j=1}^i \sum\limits_{k=1}^j 1 = \dots$
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