The sum of (n - 1) terms of the series 1 + (1 | vedclass

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(C) Let $T_k$ be the $k^{th}$ term of the series. The $k^{th}$ term is the sum of the first $k$ odd numbers,which is $T_k = k^2$.The sum of the first

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$\frac{(n - 1)n(2n - 1)}{6}$

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(C) Let $T_k$ be the $k^{th}$ term of the series. The $k^{th}$ term is the sum of the first $k$ odd numbers,which is $T_k = k^2$.The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}$.We need the sum of $(n - 1)$ terms,so we replace $n$ with $(n - 1)$ in the formula for $S_n$:$S_{n-1} = \frac{(n - 1)((n - 1) + 1)(2(n - 1) + 1)}{6}$$S_{n-1} = \frac{(n - 1)(n)(2n - 2 + 1)}{6}$$S_{n-1} = \frac{(n - 1)n(2n - 1)}{6}$.

The sum of $(n - 1)$ terms of the series $1 + (1 + 3) + (1 + 3 + 5) + \dots$ is

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