$$\sum_{r=6}^{n}{6r^2-6r}$$

asked by guest
on Nov 23, 2024 at 12:56 am



You asked:

Evaluate the expression: \(\sum_{r = 6}^{n} {6 {r}^{2} - 6 r}\)

MathBot Answer:

\[\sum_{r=6}^{n} \left(6 r^{2} - 6 r\right) = 2 n^{3} - 2 n - 240\]

$$\begin{aligned}\sum_{r=6}^{n} \left(6 r^{2} - 6 r\right) &= \sum_{r=6}^{n} 6 r^{2} + \sum_{r=6}^{n} \left(-1\right) 6 r \ \ \ \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)}}\\&= 6 \sum_{r=6}^{n} r^{2} + \sum_{r=6}^{n} \left(-1\right) 6 r \ \ \ \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} + \sum_{r=6}^{n} \left(-1\right) 6 r - 6 \sum_{r=0}^{5} r^{2} \ \ \ \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} + \sum_{r=6}^{n} \left(-1\right) 6 r - 6 \cdot \left(\frac{5}{6} + \frac{5^{2}}{2} + \frac{5^{3}}{3}\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} + \sum_{r=6}^{n} \left(-1\right) 6 r -330\\&= \sum_{r=6}^{n} \left(-1\right) 6 r + 6 \left(\frac{n}{6} + \frac{n^{2}}{2} + \frac{n^{3}}{3}\right) -330 \ \ \ \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=6}^{n} r + 6 \left(\frac{n}{6} + \frac{n^{2}}{2} + \frac{n^{3}}{3}\right) -330 \ \ \ \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 + 6 \sum_{r=0}^{5} r + 6 \left(\frac{n}{6} + \frac{n^{2}}{2} + \frac{n^{3}}{3}\right) -330 \ \ \ \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 + 6 \left(\frac{n}{6} + \frac{n^{2}}{2} + \frac{n^{3}}{3}\right) -330 + \frac{6 \cdot 5 \cdot \left(1 + 5\right)}{2} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=0}^{m} n = \frac{m(m+1)}{2}}\\&= - 6 \sum_{r=0}^{n} r + 6 \left(\frac{n}{6} + \frac{n^{2}}{2} + \frac{n^{3}}{3}\right) -330 + 90\\&= 6 \left(\frac{n}{6} + \frac{n^{2}}{2} + \frac{n^{3}}{3}\right) -330 + 90 - \frac{6 n \left(1 + n\right)}{2} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=0}^{m} n = \frac{m(m+1)}{2}}\\&= 2 n^{3} + 3 n^{2} - 3 n \left(n + 1\right) + n - 240\end{aligned}$$

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