10.1
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
\[\begin{aligned}
& \sum_{i=1}^n i^2\\
= & 2\sum_{i=2}^n {i \choose 2} + \sum_{i=1}^n {i \choose 1}\\
= & 2{n + 1\choose 3} + \frac {n (n + 1)} 2\\
= & \frac {n(n + 1)(n - 1)} 3 + \frac {n(n + 1)} 2\\
= & \frac {n(n+1)(2n+1)} 6
\end{aligned}
\]
