The sum to n terms of (2n - 1) + 2(2n - 3) + | vedclass

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(D) The $k$-th term of the series is $T_k = k(2n - (2k - 1)) = k(2n - 2k + 1) = 2nk - 2k^2 + k$.Summing from $k = 1$ to $n$:$S = \sum_{k=1}^{n} (2nk -

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$n(n + 1)(2n + 1)/6$

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(D) The $k$-th term of the series is $T_k = k(2n - (2k - 1)) = k(2n - 2k + 1) = 2nk - 2k^2 + k$.Summing from $k = 1$ to $n$:$S = \sum_{k=1}^{n} (2nk - 2k^2 + k) = 2n \sum_{k=1}^{n} k - 2 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k$.Using standard summation formulas:$S = 2n \frac{n(n+1)}{2} - 2 \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$.$S = n^2(n+1) - \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$.Factor out $\frac{n(n+1)}{6}$:$S = \frac{n(n+1)}{6} [6n - 2(2n+1) + 3] = \frac{n(n+1)}{6} [6n - 4n - 2 + 3] = \frac{n(n+1)(2n+1)}{6}$.

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

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