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\displaystyle[br]\begin{align*} \sum_{r=0}^{n} 8 - 8r &= 8 - 8(0) + \sum_{r=1}^n (8-8r) \\&= 8 + 8n - 8 \times \frac{n}{2}(n+1) \\&= 8(n+1) - 4n(n+1) \\&= 4(n+1)(2 - n) \\ & = 4(2n - n^2 +2 -n) \\& = -4n^2 +4n + 8\end{align*}
\displaystyle[br]\begin{align*} \sum_{r=0}^{n} 8 - 8r &= 8 - 8(0) + \sum_{r=1}^n (8-8r) \\&= 8 + 8n - 8 \times \frac{n}{2}(n+1) \\&= 8(n+1) - 4n(n+1) \\&= 4(n+1)(2 - n) \\ & = 4(2n - n^2 +2 -n) \\& = -4n^2 +4n + 8\end{align*}
\displaystyle[br]\begin{align*} \sum_{r=0}^{n} 8 - 8r &= 8 - 8(0) + \sum_{r=1}^n (8-8r) \\&= 8 + 8n - 8 \times \frac{n}{2}(n+1) \\&= 8(n+1) - 4n(n+1) \\&= 4(n+1)(2 - n) \\ & = 4(2n - n^2 +2 -n) \\& = -4n^2 +4n + 8\end{align*}