一些算式的证明。
\[\begin{aligned} \sum_{1\le i < j \le n} i + j &= \sum\limits_{i = 1}^{n - 1}\left(\sum\limits_{j = i + 1}^{n} j + \sum\limits_{j = i + 1}^{n} i\right) \\ &= \sum\limits_{i = 1}^{n - 1}\left[\dfrac{\left(n + i + 1\right) \times \left(n - i\right)}{2} + i\left(n - i\right)\right] \\ &= \sum\limits_{i = 1}^{n - 1}\left(\dfrac{n^2 - in + in - i^2 + n - i}{2} + in - i^2\right) \\ &= \sum\limits_{i = 1}^{n - 1}\left(\dfrac{n^2}{2} - \dfrac{3}{2}i^2 + \dfrac{n}{2} + \left(n - \dfrac{1}{2}\right)i\right) \\ &= \dfrac{n \left(n + 1\right)\left(n - 1\right)}{2} - \dfrac{n\left(n - 1\right)\left(2n - 1\right)}{4} + \dfrac{n\left(2n - 1\right)\left(n - 1\right)}{4} \\ &= \dfrac{n \left(n + 1\right)\left(n - 1\right)}{2} \end{aligned}\]\[\begin{aligned} S_2(n) &= \sum\limits_{k = 1}^{n}k^2 \\ S_2(n) &= \sum\limits_{1 \le i \le k \le n} k \\ S_2(n) &= \sum\limits_{i = 1}^{n}\sum\limits_{k = i}^{n}k \\ S_2(n) &= \sum\limits_{i = 1}^{n}\dfrac{\left(n + i\right)\left(n - i + 1\right)}{2} \\ S_2(n) &= \sum\limits_{i = 1}^{n}\dfrac{n^2 - ni + n + ni - i^2 + i}{2} \\ S_2(n) &= \sum\limits_{i = 1}^{n}\dfrac{n^2 + n - i^2 + i}{2} \\ S_2(n) &= -\dfrac{1}{2}\sum\limits_{i = 1}^{n}i^2 + \dfrac{1}{2}\sum\limits_{i = 1}^{n}\dfrac{n^2 + n + i}{2} \\ S_2(n) &= \dfrac{-S_2(n)}{2} + \sum\limits_{i = 1}^{n}\dfrac{n^2 + n + i}{2} \\ \dfrac{3S_2(n)}{2} &= \sum\limits_{i = 1}^{n}\dfrac{n^2 + n + i}{2} \\ S_2(n) &= \dfrac{1}{3}\sum\limits_{i = 1}^{n}\left(n^2 + n + i\right) \\ S_2(n) &= \dfrac{n^3 + n^2}{3} + \dfrac{n \left(n + 1\right)}{6} \\ S_2(n) &= \dfrac{2n^2 \left(n + 1\right)}{6} + \dfrac{n \left(n + 1\right)}{6} \\ S_2(n) &= \dfrac{n\left[2n \left(n + 1\right) + n + 1\right]}{6} \\ S_2(n) &= \dfrac{n\left(2n^2 + n + 1\right)}{6} \\ S_2(n) &= \dfrac{n\left(n + 1\right)\left(2n + 1\right)}{6} \\ \end{aligned}\] 标签：right,15,limits,闲话,sum,dfrac,2n,2023.8,left From： https://www.cnblogs.com/User-Unauthorized/p/17631961.html