Find the sum to n terms of the series 3 times | vedclass

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Question

(A) The $n^{th}$ term of the series is given by $a_{n} = (2n+1)n^{2} = 2n^{3} + n^{2}$.Sum to $n$ terms $S_{n} = \sum_{k=1}^{n} a_{k} = \sum_{k=1}^{n}

Vedclass · Accepted Answer

$\frac{n(n+1)(3n^{2}+5n+1)}{6}$

Vedclass · Answer

(A) The $n^{th}$ term of the series is given by $a_{n} = (2n+1)n^{2} = 2n^{3} + n^{2}$.Sum to $n$ terms $S_{n} = \sum_{k=1}^{n} a_{k} = \sum_{k=1}^{n} (2k^{3} + k^{2})$.$S_{n} = 2 \sum_{k=1}^{n} k^{3} + \sum_{k=1}^{n} k^{2}$.Using the formulas $\sum k^{3} = [\frac{n(n+1)}{2}]^{2}$ and $\sum k^{2} = \frac{n(n+1)(2n+1)}{6}$:$S_{n} = 2 [\frac{n(n+1)}{2}]^{2} + \frac{n(n+1)(2n+1)}{6}$.$S_{n} = \frac{n^{2}(n+1)^{2}}{2} + \frac{n(n+1)(2n+1)}{6}$.$S_{n} = \frac{n(n+1)}{2} [n(n+1) + \frac{2n+1}{3}]$.$S_{n} = \frac{n(n+1)}{2} [\frac{3n^{2} + 3n + 2n + 1}{3}]$.$S_{n} = \frac{n(n+1)(3n^{2} + 5n + 1)}{6}$.

Find the sum to $n$ terms of the series $3 \times 1^{2} + 5 \times 2^{2} + 7 \times 3^{2} + \dots$

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