1*2+2*3+3*4+4*5+...+2024*2025
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MathBot Answer:
The sum of the sequence is \(\displaystyle 2767921200\)
This is an arithmetic sequence.
The nth term in this sequence is given by the formula:
Explicit Formula: $$a_n=n \left(n + 1\right)$$
Recursive Formula: $$a_n=a_{n-1} + 2 n, \text{where } a_1=1 \cdot 2$$
Summation Formula:
$$\begin{aligned} S_n&=\sum_{i=1}^{n} a_{i} \\ &=\sum_{i=1}^{n} i \left(i + 1\right) \\ &= \sum_{i=0}^{n - 1} \left(i + 1\right) \left(i + 2\right) \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k + p}^{m + p} f{\left(n - p \right)} = \sum_{n=k}^{m} f{\left(n \right)}}\\&= \sum_{i=0}^{n - 1} \left(i^{2} + 3 i + 2\right)\\&= \sum_{i=0}^{n - 1} 3 i + \sum_{i=0}^{n - 1} \left(i^{2} + 2\right) \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k}^{m} \left(f{\left(n \right)} + g{\left(n \right)}\right) = \sum_{n=k}^{m} f{\left(n \right)} + \sum_{n=k}^{m} g{\left(n \right)}}\\&= 3 \sum_{i=0}^{n - 1} i + \sum_{i=0}^{n - 1} \left(i^{2} + 2\right) \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k}^{m} C f{\left(n \right)} = C \sum_{n=k}^{m} f{\left(n \right)}}\\&= \sum_{i=0}^{n - 1} \left(i^{2} + 2\right) + \frac{3 \left(n - 1\right) \left(1 + n - 1\right)}{2} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=0}^{m} n = \frac{m(m+1)}{2}}\\&= \sum_{i=0}^{n - 1} 2 + \sum_{i=0}^{n - 1} i^{2} + \frac{3 \left(n - 1\right) \left(1 + n - 1\right)}{2} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k}^{m} \left(f{\left(n \right)} + g{\left(n \right)}\right) = \sum_{n=k}^{m} f{\left(n \right)} + \sum_{n=k}^{m} g{\left(n \right)}}\\&= \sum_{i=0}^{n - 1} i^{2} + 2 n + \frac{3 \left(n - 1\right) \left(1 + n - 1\right)}{2} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=0}^{m} C = C(m+1)}\\&= 2 n + \frac{3 \left(n - 1\right) \left(1 + n - 1\right)}{2} + \frac{\left(n - 1\right)}{6} + \frac{\left(n - 1\right)^{2}}{2} + \frac{\left(n - 1\right)^{3}}{3} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=0}^{m} n^2 = \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}}\\&=\frac{n \left(n^{2} + 3 n + 2\right)}{3} \end{aligned}$$where $n$ is the is the term number.
Use the explicit formula to find the index of the final term in the sequence.$$\begin{aligned}2024 \cdot 2025 &= n \left(n + 1\right) \\ n &= 2024\end{aligned}$$Then use the summation formula to solve.$$\begin{aligned}S_n &= \frac{n \left(n^{2} + 3 n + 2\right)}{3} \\S_{2024} &= \frac{2024 \cdot \left(2 + 3 \cdot 2024 + 2024^{2}\right)}{3}\\S_{2024} &= 2767921200\end{aligned}$$