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The sum of $(n-1)$ terms of $1+(1+3)+(1+3+5)+\ldots \ldots$ is
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Verified Answer
The correct answer is:
$\frac{n(n-1)(2 n-1)}{6}$
Let $T_n$ be the $n^{\text {th }}$ term of the series $T_n=2 \sum n-\sum 1$
$\Rightarrow \quad T_n=\frac{2 n(n+1)}{2}-n=n^2$
$\therefore S_n=\sum_{k=1}^n\left(k^2\right)=\frac{n(n+1)(2 n+1)}{6}$
Hence sum of $(n-1)$ terms $S_{n-1}=\frac{(n-1) n(2 n-1)}{6}$
$\Rightarrow \quad T_n=\frac{2 n(n+1)}{2}-n=n^2$
$\therefore S_n=\sum_{k=1}^n\left(k^2\right)=\frac{n(n+1)(2 n+1)}{6}$
Hence sum of $(n-1)$ terms $S_{n-1}=\frac{(n-1) n(2 n-1)}{6}$
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