The sum to n terms of the series 1^2 + (1^2 + | vedclass

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(A) The $r$-th term of the series is $t_r = 1^2 + 3^2 + 5^2 + \dots + (2r - 1)^2$.$t_r = \sum_{k=1}^r (2k - 1)^2 = \sum_{k=1}^r (4k^2 - 4k + 1)$.Using

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

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(A) The $r$-th term of the series is $t_r = 1^2 + 3^2 + 5^2 + \dots + (2r - 1)^2$.$t_r = \sum_{k=1}^r (2k - 1)^2 = \sum_{k=1}^r (4k^2 - 4k + 1)$.Using standard summation formulas:$t_r = 4 \cdot \frac{r(r+1)(2r+1)}{6} - 4 \cdot \frac{r(r+1)}{2} + r$.Simplifying this expression:$t_r = \frac{2r(r+1)(2r+1)}{3} - 2r(r+1) + r = \frac{2r(2r^2 + 3r + 1) - 6r^2 - 6r + 3r}{3} = \frac{4r^3 + 6r^2 + 2r - 6r^2 - 3r}{3} = \frac{4r^3 - r}{3}$.The sum of $n$ terms is $S_n = \sum_{r=1}^n t_r = \sum_{r=1}^n \frac{4r^3 - r}{3} = \frac{4}{3} \sum_{r=1}^n r^3 - \frac{1}{3} \sum_{r=1}^n r$.$S_n = \frac{4}{3} \left[ \frac{n(n+1)}{2} \right]^2 - \frac{1}{3} \left[ \frac{n(n+1)}{2} \right]$.$S_n = \frac{n(n+1)}{6} [2n(n+1) - 1] = \frac{1}{6} n(n+1)(2n^2 + 2n - 1)$.

The sum to $n$ terms of the series $1^2 + (1^2 + 3^2) + (1^2 + 3^2 + 5^2) + \dots$ is

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