$$\sum_{r=6}^{n}{6r^2-6}$$
You asked:
MathBot Answer:
\[\sum_{r=6}^{n} \left(6 r^{2} - 6\right) = 2 n^{3} + 3 n^{2} - 5 n - 300\]
$$\begin{aligned}\sum_{r=6}^{n} \left(6 r^{2} - 6\right) &= \sum_{r=6}^{n} -6 + \sum_{r=6}^{n} 6 r^{2} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k}^{m} \left(f{\left(n \right)} + g{\left(n \right)}\right) = \sum_{n=k}^{m} f{\left(n \right)} + \sum_{n=k}^{m} g{\left(n \right)}}\\&= \sum_{r=0}^{n} -6 + \sum_{r=6}^{n} 6 r^{2} - \sum_{r=0}^{5} -6 \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k}^{m} f{\left(n \right)} = \sum_{n=0}^{m} f{\left(n \right)} - \sum_{n=0}^{k-1} f{\left(n \right)}}\\&= \sum_{r=0}^{n} -6 + \sum_{r=6}^{n} 6 r^{2} - 6 \left(-6\right) \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=0}^{m} C = C(m+1)}\\&= \sum_{r=0}^{n} -6 + \sum_{r=6}^{n} 6 r^{2} + 36\\&= \sum_{r=6}^{n} 6 r^{2} + 36 + \left(n + 1\right) \left(-6\right) \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=0}^{m} C = C(m+1)}\\&= 6 \sum_{r=6}^{n} r^{2} + 36 + \left(n + 1\right) \left(-6\right) \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k}^{m} C f{\left(n \right)} = C \sum_{n=k}^{m} f{\left(n \right)}}\\&= 6 \sum_{r=0}^{n} r^{2} - 6 \sum_{r=0}^{5} r^{2} + 36 + \left(n + 1\right) \left(-6\right) \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k}^{m} f{\left(n \right)} = \sum_{n=0}^{m} f{\left(n \right)} - \sum_{n=0}^{k-1} f{\left(n \right)}}\\&= 6 \sum_{r=0}^{n} r^{2} - 6 \cdot \left(\frac{5}{6} + \frac{5^{2}}{2} + \frac{5^{3}}{3}\right) + 36 + \left(n + 1\right) \left(-6\right) \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=0}^{m} n^2 = \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}}\\&= 6 \sum_{r=0}^{n} r^{2} -330 + 36 + \left(n + 1\right) \left(-6\right)\\&= 6 \left(\frac{n}{6} + \frac{n^{2}}{2} + \frac{n^{3}}{3}\right) -330 + 36 + \left(n + 1\right) \left(-6\right) \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=0}^{m} n^2 = \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}}\\&= 2 n^{3} + 3 n^{2} - 5 n - 300\end{aligned}$$