Class 11MathematicsSequences and Seriesnth term of special series, Sum to n terms and Infinite number of terms
Difficult$\sum_{k=1}^5 \frac{1^3+2^3+\ldots+k^3}{1+3+5+\ldots+(2 k-1)}$ is equal to (in $.5$)
- A$22$
- B$24$
- C$28$
- D$32$
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Let $S_{k} = \frac{1+2+\ldots+k}{k}$ and $\sum_{j=1}^n S_j^2 = \frac{n}{A}(Bn^2 + Cn + D)$,where $A, B, C, D \in \mathbb{N}$ and $A$ has the least value. Then:
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View SolutionIf $1^4+2^4+3^4+\ldots+n^4=f(n) \left(1^2+2^2+\ldots+n^2\right)$,for all $n \in N$,then $f(4)$ is equal to
If $2^3+4^3+6^3+\ldots+(2n)^3 = h n^2(n+1)^2$,then $h$ is equal to
DifficultAP EAMCET 2001
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