- \[(a_1 + a_2 + a_3 + ... + a_n) ^ 2 = a_1^2 + a_2^2 + ... + a_n^2 + 2a_1a_2 + 2a_1a_3 + ... 2a_{n-1}a_n \]
- \[\sum_{k = 1}^{n}k ^ 2 = \frac{n(n + 1)(2n + 1)}{6} \]
证明：
\[(n + 1) ^ 3 = n ^ 3 + 3n ^ 2 + 3n + 1 \]\[(n + 1) ^ 3 - n ^ 3 = 3n ^ 2 + 3n + 1 \]试着写几项：
\[2^3 - 1^3 = 3 \times 1^3 + 3 \times 1 + 1 \]\[3^3 - 2^3 = 3 \times 2^3 + 3 \times 2 + 1 \]\[...... \]\[(n + 1) ^ 3 - n ^ 3 = 3n ^ 2 + 3n + 1 \]累加得：
\[\begin{aligned} (n + 1) ^ 3 - 1 &= 3 \times \sum_{k = 1}^{n}k ^ 2 + 3 \sum_{k = 1}^{n}k + n\\ &= 3 \times \sum_{k = 1}^{n}k ^ 2 + 3 \times \frac{n(1 + n)}{2} + n \end{aligned} \]浅移项化简一下得：
\[\sum_{k = 1}^{n}k ^ 2 = \frac{n(n + 1)(2n + 1)}{6} \] 标签：...,frac,sum,知识,times,数学,3n,简单,2a From： https://www.cnblogs.com/yduck/p/17608419.html