Find the sum to n terms of the series whose n | vedclass

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(A) The $n^{th}$ term is $a_n = (2n-1)^2 = 4n^2 - 4n + 1$. The sum to $n$ terms is $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (4k^2 - 4k + 1)$. Using the

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$\frac{n(2n-1)(2n+1)}{3}$

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(A) The $n^{th}$ term is $a_n = (2n-1)^2 = 4n^2 - 4n + 1$. The sum to $n$ terms is $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (4k^2 - 4k + 1)$. Using the summation formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$,$\sum k = \frac{n(n+1)}{2}$,and $\sum 1 = n$: $S_n = 4 \left[ \frac{n(n+1)(2n+1)}{6} \right] - 4 \left[ \frac{n(n+1)}{2} \right] + n$ $S_n = \frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n$ $S_n = n \left[ \frac{2(2n^2+3n+1) - 6(n+1) + 3}{3} \right]$ $S_n = n \left[ \frac{4n^2 + 6n + 2 - 6n - 6 + 3}{3} \right]$ $S_n = n \left[ \frac{4n^2 - 1}{3} \right] = \frac{n(2n-1)(2n+1)}{3}$.

Find the sum to $n$ terms of the series whose $n^{th}$ term is given by $(2n-1)^2$.

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