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

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Question

(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: $T_k = 1 + 3 + 5 + \dots + (2k - 1) = k^2$

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

Vedclass · Answer

(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: $T_k = 1 + 3 + 5 + \dots + (2k - 1) = k^2$.We need to find the sum of $(n - 1)$ terms,which is $S_{n-1} = \sum_{k=1}^{n-1} T_k = \sum_{k=1}^{n-1} k^2$.Using the formula for the sum of squares of the first $m$ natural numbers,$\sum_{k=1}^{m} k^2 = \frac{m(m + 1)(2m + 1)}{6}$.Substituting $m = n - 1$:$S_{n-1} = \frac{(n - 1)((n - 1) + 1)(2(n - 1) + 1)}{6} = \frac{(n - 1)(n)(2n - 2 + 1)}{6} = \frac{n(n - 1)(2n - 1)}{6}$.

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

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