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If the $n^{\text {th }}$ term of a series be $3+n(n-1)$, then the sum of $n$ terms of the series is
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Verified Answer
The correct answer is:
$\frac{n^3+8 n}{3}$
Here, $T_n=3+n(n-1)=3+n^2-n$
Now sum $S=\Sigma T_n=\Sigma\left(3+n^2-n\right)$
$=3 n+\frac{1}{6} n(n+1)(2 n+1)-\frac{n(n+1)}{2}$
$=\frac{1}{6} n(n+1)[2 n+1-3]+3 n=\frac{n^3+8 n}{3}$
Now sum $S=\Sigma T_n=\Sigma\left(3+n^2-n\right)$
$=3 n+\frac{1}{6} n(n+1)(2 n+1)-\frac{n(n+1)}{2}$
$=\frac{1}{6} n(n+1)[2 n+1-3]+3 n=\frac{n^3+8 n}{3}$
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