$$\sum_{i=1}^{n}{i(i+1)/2}$$

asked by guest
on Nov 29, 2024 at 12:55 pm



You asked:

Evaluate the expression: \(\sum_{i = 1}^{n} {\frac{i \cdot \left(i + 1\right)}{2}}\)

MathBot Answer:

\[\sum_{i=1}^{n} \frac{i \left(i + 1\right)}{2} = \frac{n \left(n^{2} + 3 n + 2\right)}{6}\]

$$\begin{aligned}\sum_{i=1}^{n} \frac{i \left(i + 1\right)}{2} &= \frac{\sum_{i=1}^{n} i \left(i + 1\right)}{2} \ \ \ \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)}}\\&= \frac{\sum_{i=0}^{n - 1} \left(i + 1\right) \left(i + 2\right)}{2} \ \ \ \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)}}\\&= \frac{\sum_{i=0}^{n - 1} \left(i^{2} + 3 i + 2\right)}{2}\\&= \frac{\sum_{i=0}^{n - 1} 3 i}{2} + \frac{\sum_{i=0}^{n - 1} \left(i^{2} + 2\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)}}\\&= \frac{3 \sum_{i=0}^{n - 1} i}{2} + \frac{\sum_{i=0}^{n - 1} \left(i^{2} + 2\right)}{2} \ \ \ \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)}}\\&= \frac{\sum_{i=0}^{n - 1} \left(i^{2} + 2\right)}{2} + \frac{\frac{3}{2} \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}}\\&= \frac{\sum_{i=0}^{n - 1} 2}{2} + \frac{\sum_{i=0}^{n - 1} i^{2}}{2} + \frac{\frac{3}{2} \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)}}\\&= \frac{\sum_{i=0}^{n - 1} i^{2}}{2} + \frac{n 2}{2} + \frac{\frac{3}{2} \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)}\\&= \frac{\frac{\left(n - 1\right)}{6} + \frac{\left(n - 1\right)^{2}}{2} + \frac{\left(n - 1\right)^{3}}{3}}{2} + \frac{n 2}{2} + \frac{\frac{3}{2} \left(n - 1\right) \left(1 + n - 1\right)}{2} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=0}^{m} n^2 = \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}}\\&= \frac{3 n \left(n - 1\right)}{4} + \frac{13 n}{12} + \frac{\left(n - 1\right)^{3}}{6} + \frac{\left(n - 1\right)^{2}}{4} - \frac{1}{12}\end{aligned}$$

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