观察以下式子：
\[\begin{aligned} &1^3 = 1 = 1\\ &2^3 = 8 = 3 + 5\\ &3^3 = 27 = 7 + 9 + 11 \end{aligned} \]猜到：
\[n^3 = \frac 1 2 n[(n^2 - n + 1) + (n^2 -n+1+2n-2)]\\ \]可证明正确。
那么
\[\begin{aligned} &\sum_{i=1}^n i^3\\ = & \frac 1 2 \times \frac{n (n + 1)} 2 \times (1 + n^2 + n - 1)\\ = & \frac {n^2(n+1)^2} 4 \end{aligned} \]根据
\[n^2 = 2{n \choose 2} + {n \choose 1} \]可知
\[\begin{aligned} & \sum_{i=1}^n i^2\\ = & 2\sum_{i=2}^n {i \choose 2} + \sum_{i=1}^n {i \choose 1}\\ = & 2{n + 1\choose 3} + \frac {n (n + 1)} 2\\ = & \frac {n(n + 1)(n - 1)} 3 + \frac {n(n + 1)} 2\\ = & \frac {n(n+1)(2n+1)} 6 \end{aligned} \] 标签：10.1,begin,frac,sum,choose,end,aligned From： https://www.cnblogs.com/Estelle-N/p/18444271