If the n^{th} term of a series is 3 + n(n - 1 | vedclass

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(B) Given the $n^{th}$ term $T_n = 3 + n(n - 1) = n^2 - n + 3$.The sum of $n$ terms $S_n$ is given by $\sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 - k +

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$\frac{n^3 + 8n}{3}$

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(B) Given the $n^{th}$ term $T_n = 3 + n(n - 1) = n^2 - n + 3$.The sum of $n$ terms $S_n$ is given by $\sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 - k + 3)$.Using the standard summation formulas:$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$,and $\sum_{k=1}^{n} 3 = 3n$.$S_n = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} + 3n$.Taking $\frac{n}{6}$ as a common factor:$S_n = \frac{n}{6} [(n+1)(2n+1) - 3(n+1) + 18]$.$S_n = \frac{n}{6} [2n^2 + 3n + 1 - 3n - 3 + 18] = \frac{n}{6} [2n^2 + 16]$.$S_n = \frac{2n(n^2 + 8)}{6} = \frac{n(n^2 + 8)}{3} = \frac{n^3 + 8n}{3}$.

If the $n^{th}$ term of a series is $3 + n(n - 1)$,then the sum of $n$ terms of the series is

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