Class 11MathematicsSequences and Seriesnth term of special series, Sum to n terms and Infinite number of terms
EasyIf $1^4+2^4+3^4+\ldots+n^4=f(n) \left(1^2+2^2+\ldots+n^2\right)$,for all $n \in N$,then $f(4)$ is equal to
- A$\frac{58}{5}$
- B$\frac{57}{5}$
- C$\frac{59}{5}$
- D$\frac{56}{5}$
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Similar Questions
The sum of all terms of the $n^{th}$ bracket of the sequence $(1), (3, 5), (7, 9, 11), \dots$ is equal to:
Let $a_{1}=b_{1}=1$,$a_{n}=a_{n-1}+2$,and $b_{n}=a_{n}+b_{n-1}$ for every natural number $n \geq 2$. Then $\sum_{n=1}^{15} a_{n} \cdot b_{n}$ is equal to $.........$
$\sum_{n=1}^5 n(n^2+n+1) = $
The $20^{th}$ term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + \dots$ will be
Easy
View SolutionThe sum of $(1^2-1+1)(1!) + (2^2-2+1)(2!) + \ldots + (n^2-n+1)(n!)$ is
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