Math Problem Solver

\[\sum_{r=1}^{n} \left(r^{3} + r\right) = \frac{n \left(n^{3} + 2 n^{2} + 3 n + 2\right)}{4}\]

$$\begin{aligned}\sum_{r=1}^{n} \left(r^{3} + r\right) &= \sum_{r=1}^{n} r^{3} + \sum_{r=1}^{n} 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)}}\\&= \sum_{r=1}^{n} r + \sum_{r=0}^{n - 1} \left(r + 1\right)^{3} \ \ \ \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)}}\\&= \sum_{r=1}^{n} r + \sum_{r=0}^{n - 1} \left(r^{3} + 3 r^{2} + 3 r + 1\right)\\&= \sum_{r=0}^{n - 1} 3 r^{2} + \sum_{r=1}^{n} r + \sum_{r=0}^{n - 1} \left(r^{3} + 3 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)}}\\&= 3 \sum_{r=0}^{n - 1} r^{2} + \sum_{r=1}^{n} r + \sum_{r=0}^{n - 1} \left(r^{3} + 3 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)}}\\&= \sum_{r=1}^{n} r + \sum_{r=0}^{n - 1} \left(r^{3} + 3 r + 1\right) + 3 \left(\frac{n - 1}{6} + \frac{\left(n - 1\right)^{2}}{2} + \frac{\left(n - 1\right)^{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}}\\&= \sum_{r=0}^{n - 1} 3 r + \sum_{r=1}^{n} r + \sum_{r=0}^{n - 1} \left(r^{3} + 1\right) + 3 \left(\frac{n - 1}{6} + \frac{\left(n - 1\right)^{2}}{2} + \frac{\left(n - 1\right)^{3}}{3}\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)}}\\&= 3 \sum_{r=0}^{n - 1} r + \sum_{r=1}^{n} r + \sum_{r=0}^{n - 1} \left(r^{3} + 1\right) + 3 \left(\frac{n - 1}{6} + \frac{\left(n - 1\right)^{2}}{2} + \frac{\left(n - 1\right)^{3}}{3}\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)}}\\&= \sum_{r=1}^{n} r + \sum_{r=0}^{n - 1} \left(r^{3} + 1\right) + 3 \left(\frac{n - 1}{6} + \frac{\left(n - 1\right)^{2}}{2} + \frac{\left(n - 1\right)^{3}}{3}\right) + \frac{3 \left(n - 1\right) \left(1 + n - 1\right)}{2} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=0}^{m} n = \frac{m(m+1)}{2}}\\&= \sum_{r=0}^{n - 1} 1 + \sum_{r=0}^{n - 1} r^{3} + \sum_{r=1}^{n} r + 3 \left(\frac{n - 1}{6} + \frac{\left(n - 1\right)^{2}}{2} + \frac{\left(n - 1\right)^{3}}{3}\right) + \frac{3 \left(n - 1\right) \left(1 + n - 1\right)}{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 - 1} r^{3} + \sum_{r=1}^{n} r + 3 \left(\frac{n - 1}{6} + \frac{\left(n - 1\right)^{2}}{2} + \frac{\left(n - 1\right)^{3}}{3}\right) + 1 n + \frac{3 \left(n - 1\right) \left(1 + n - 1\right)}{2} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=0}^{m} C = C(m+1)}\\&= \sum_{r=1}^{n} r + 3 \left(\frac{n - 1}{6} + \frac{\left(n - 1\right)^{2}}{2} + \frac{\left(n - 1\right)^{3}}{3}\right) + 1 n + \frac{3 \left(n - 1\right) \left(1 + n - 1\right)}{2} + \frac{\left(n - 1\right)^{2}}{4} + \frac{\left(n - 1\right)^{3}}{2} + \frac{\left(n - 1\right)^{4}}{4} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=0}^{m} n^3 = \frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4}}\\&= 3 \left(\frac{n - 1}{6} + \frac{\left(n - 1\right)^{2}}{2} + \frac{\left(n - 1\right)^{3}}{3}\right) + 1 n + \frac{n \left(1 + n\right)}{2} + \frac{3 \left(n - 1\right) \left(1 + n - 1\right)}{2} + \frac{\left(n - 1\right)^{2}}{4} + \frac{\left(n - 1\right)^{3}}{2} + \frac{\left(n - 1\right)^{4}}{4} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=1}^{m} n = \frac{m(m+1)}{2}}\\&= \frac{n \left(n^{3} + 2 n^{2} + 3 n + 2\right)}{4}\end{aligned}$$