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

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Question

(A) The $r$-th term of the series is $t_r = 1^2 + 3^2 + 5^2 + ... + (2r - 1)^2$.Using the formula for the sum of squares of the first $r$ odd numbers,

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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 $t_r = 1^2 + 3^2 + 5^2 + ... + (2r - 1)^2$.Using the formula for the sum of squares of the first $r$ odd numbers,$t_r = \sum_{k=1}^r (2k - 1)^2 = \sum_{k=1}^r (4k^2 - 4k + 1)$.$t_r = 4 \sum_{k=1}^r k^2 - 4 \sum_{k=1}^r k + \sum_{k=1}^r 1 = 4 \cdot \frac{r(r+1)(2r+1)}{6} - 4 \cdot \frac{r(r+1)}{2} + r$.$t_r = \frac{2r(r+1)(2r+1)}{3} - 2r(r+1) + r = \frac{r}{3} [2(2r^2 + 3r + 1) - 6(r+1) + 3] = \frac{r}{3} [4r^2 + 6r + 2 - 6r - 6 + 3] = \frac{r(4r^2 - 1)}{3} = \frac{4r^3 - r}{3}$.Now,the sum of $n$ terms is $S_n = \sum_{r=1}^n t_r = \sum_{r=1}^n \frac{4r^3 - r}{3} = \frac{4}{3} \sum_{r=1}^n r^3 - \frac{1}{3} \sum_{r=1}^n r$.$S_n = \frac{4}{3} \cdot \frac{n^2(n+1)^2}{4} - \frac{1}{3} \cdot \frac{n(n+1)}{2} = \frac{n^2(n+1)^2}{3} - \frac{n(n+1)}{6}$.$S_n = \frac{n(n+1)}{6} [2n(n+1) - 1] = \frac{n(n+1)(2n^2 + 2n - 1)}{6}$.

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

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