\[(a_1 + a_2 + a_3 + ... + a_n) ^ 2 = a_1^2 + a_2^2 + ... + a_n^2 + 2a_1a_2 + 2a_1a_3 + ... 2a_{n-1}a_n \]
\[\sum_{k = 1}^{n}k ^ 2 = \frac{n(n + 1)(2n + 1)}{6} \]

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：

\[\sum_{k = 1}^{n}k ^ 2 = \frac{n(n + 1)(2n + 1)}{6} \]

\[a^{n + 1} - 1 = (a - 1)(1 + a + a^2+...+a^n) \]