Evaluate $$\sum_{r=10}^{25}{2r(r+1)}$$
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MathBot Answer:
\[\sum_{r=10}^{25} 2 r \left(r + 1\right) = 11040\]
$$\begin{aligned}\sum_{r=10}^{25} 2 r \left(r + 1\right) &= 2 \sum_{r=10}^{25} r \left(r + 1\right) \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k}^{m} C f{\left(n \right)} = C \sum_{n=k}^{m} f{\left(n \right)}}\\&= - 2 \sum_{r=0}^{9} r \left(r + 1\right) + 2 \sum_{r=0}^{25} r \left(r + 1\right) \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k}^{m} f{\left(n \right)} = \sum_{n=0}^{m} f{\left(n \right)} - \sum_{n=0}^{k-1} f{\left(n \right)}}\\&= - 2 \sum_{r=1}^{10} r \left(r - 1\right) + 2 \sum_{r=0}^{25} r \left(r + 1\right) \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k + p}^{m + p} f{\left(n - p \right)} = \sum_{n=k}^{m} f{\left(n \right)}}\\&= - 2 \sum_{r=1}^{10} \left(r^{2} - r\right) + 2 \sum_{r=0}^{25} r \left(r + 1\right)\\&= - 2 \sum_{r=1}^{10} r^{2} - 2 \sum_{r=1}^{10} - r + 2 \sum_{r=0}^{25} r \left(r + 1\right) \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k}^{m} \left(f{\left(n \right)} + g{\left(n \right)}\right) = \sum_{n=k}^{m} f{\left(n \right)} + \sum_{n=k}^{m} g{\left(n \right)}}\\&= - 2 \sum_{r=1}^{10} - r + 2 \sum_{r=0}^{25} r \left(r + 1\right) - 2 \cdot \left(\frac{10}{6} + \frac{10^{2}}{2} + \frac{10^{3}}{3}\right) \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=1}^{m} n^2 = \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}}\\&= - 2 \sum_{r=1}^{10} - r + 2 \sum_{r=0}^{25} r \left(r + 1\right) -770\\&= 2 \sum_{r=1}^{10} r + 2 \sum_{r=0}^{25} r \left(r + 1\right) -770 \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k}^{m} C f{\left(n \right)} = C \sum_{n=k}^{m} f{\left(n \right)}}\\&= 2 \sum_{r=0}^{25} r \left(r + 1\right) -770 + \frac{2 \cdot 10 \cdot \left(1 + 10\right)}{2} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=1}^{m} n = \frac{m(m+1)}{2}}\\&= 2 \sum_{r=0}^{25} r \left(r + 1\right) -770 + 110\\&= 2 \sum_{r=1}^{26} r \left(r - 1\right) -770 + 110 \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k + p}^{m + p} f{\left(n - p \right)} = \sum_{n=k}^{m} f{\left(n \right)}}\\&= 2 \sum_{r=1}^{26} \left(r^{2} - r\right) -770 + 110\\&= 2 \sum_{r=1}^{26} - r + 2 \sum_{r=1}^{26} r^{2} -770 + 110 \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k}^{m} \left(f{\left(n \right)} + g{\left(n \right)}\right) = \sum_{n=k}^{m} f{\left(n \right)} + \sum_{n=k}^{m} g{\left(n \right)}}\\&= 2 \sum_{r=1}^{26} - r + 2 \cdot \left(\frac{26}{6} + \frac{26^{2}}{2} + \frac{26^{3}}{3}\right) -770 + 110 \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=1}^{m} n^2 = \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}}\\&= 2 \sum_{r=1}^{26} - r -770 + 110 + 12402\\&= - 2 \sum_{r=1}^{26} r -770 + 110 + 12402 \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k}^{m} C f{\left(n \right)} = C \sum_{n=k}^{m} f{\left(n \right)}}\\&= -770 - \frac{2 \cdot 26 \cdot \left(1 + 26\right)}{2} + 110 + 12402 \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=1}^{m} n = \frac{m(m+1)}{2}}\\&= -770 -702 + 110 + 12402\\&= 11040\end{aligned}$$