For any integer n geq 1,the sum sum_{k=1}^n k | vedclass

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(C) We need to evaluate the sum $S = \sum_{k=1}^n k(k+2)$.Expanding the term inside the summation,we get $k^2 + 2k$.Thus,$S = \sum_{k=1}^n (k^2 + 2k)

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$\frac{n(n+1)(2n+7)}{6}$

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(C) We need to evaluate the sum $S = \sum_{k=1}^n k(k+2)$.Expanding the term inside the summation,we get $k^2 + 2k$.Thus,$S = \sum_{k=1}^n (k^2 + 2k) = \sum_{k=1}^n k^2 + 2 \sum_{k=1}^n k$.Using the standard summation formulas $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^n k = \frac{n(n+1)}{2}$,we have:$S = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2}$.$S = \frac{n(n+1)(2n+1)}{6} + n(n+1)$.Factoring out $\frac{n(n+1)}{6}$,we get:$S = \frac{n(n+1)}{6} [ (2n+1) + 6 ]$.$S = \frac{n(n+1)(2n+7)}{6}$.

For any integer $n \geq 1$,the sum $\sum_{k=1}^n k(k+2)$ is equal to

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