考虑一个问题
$$1 \leq n \leq 1e7,求\sum_{1 \leq i< j \leq n}(a_{i}b_{j}-a_{j}b_{i})^{2}(mod\quad1e9+7)$$
结论——拉格朗日恒等式
\[(\sum_{i=1}^{n}a_{i}^{2})(\sum_{i=1}^{n}b_{i}^{2})=(\sum_{i=1}^{n}a_{i}b_{i})^{2}+\sum_{1 \leq i< j \leq n}(a_{i}b_{j}-a_{j}b_{i})^{2}
\]
拉格朗日恒等式的证明
法1 directly
证明：
\[左边=\sum_{i=1}^{n}a_{i}^{2}b_{i}^{2}+\sum_{1 \leq i< j \leq n}a_{i}^{2}b_{j}^{2}+\sum_{1 \leq i< j \leq n}a_{j}^{2}b_{i}^{2}
\]
\[右边=\sum_{i=1}^{n}a_{i}^{2}b_{i}^{2}+2\sum_{1 \leq i< j \leq n}a_{i}b_{j}a_{j}b_{i}+\sum_{1 \leq i< j \leq n}a_{i}^{2}b_{j}^{2}+\sum_{1 \leq i< j \leq n}a_{j}^{2}b_{i}^{2}-2\sum_{1 \leq i< j \leq n}a_{i}b_{j}a_{j}b_{i}
\]
左边=右边
证毕#
法2 数学归纳法
证明：
\[1^{\circ}\quad当i=1时，左边=a_{1}^{2}b_{1}^{2}=右边，满足原式
\]
\[2^{\circ}\quad假设当i=n-1时成立，则有(\sum_{i=1}^{n-1}a_{i}^{2})(\sum_{i=1}^{n-1}b_{i}^{2})=(\sum_{i=1}^{n-1}a_{i}b_{i})^{2}+\sum_{1 \leq i< j \leq n-1}(a_{i}b_{j}-a_{j}b_{i})^{2}
\]
\[3^{\circ}\quad当i=n时，(\sum_{i=1}^{n}a_{i}^{2})(\sum_{i=1}^{n}b_{i}^{2})=(\sum_{i=1}^{n-1}a_{i}^{2})(\sum_{i=1}^{n-1}b_{i}^{2})+a_{n}^{2}+b_{n}^{2}+\sum_{i=1}^{n-1}a_{i}^{2}b_{n}^{2}+\sum_{i=1}^{n-1}b_{i}^{2}a_{n}^{2}
\]
\[(\sum_{i=1}^{n}a_{i}b_{i})^{2}+\sum_{1 \leq i< j \leq n}(a_{i}b_{j}-a_{j}b_{i})^{2}=(\sum_{i=1}^{n-1}a_{i}b_{i})^{2}+2a_{n}b_{n}\sum_{i=1}^{n-1}a_{i}b_{i}+a_{n}^{2}+b_{n}^{2}+\sum_{1 \leq i< j \leq n-1}(a_{i}b_{j}-a_{j}b_{i})^{2}-2a_{n}b_{n}\sum_{i=1}^{n-1}a_{i}b_{i}
\]
\[可得(\sum_{i=1}^{n}a_{i}^{2})(\sum_{i=1}^{n}b_{i}^{2})=(\sum_{i=1}^{n}a_{i}b_{i})^{2}+\sum_{1 \leq i< j \leq n}(a_{i}b_{j}-a_{j}b_{i})^{2}
\]
证毕#
拉格朗日恒等式
原文：https://www.cnblogs.com/re0acm/p/15208368.html
