考慮一個問題
$$1 \leq n \leq 1e7,求\sum_{1 \leq i< j \leq n}(a_{i}b_{j}-a_{j}b_{i})^{2}(mod\quad1e9+7)$$結論——拉格朗日恆等式
\[(\sum_{i=1}^{n}a_{i}^{2})(\sum_{i=1}^{n}b_{i}^{2})=(\sum_{i=1}^{n}a_{i}b_{i})^{2}+\sum_{1 \leq i< j \leq n}(a_{i}b_{j}-a_{j}b_{i})^{2} \]
拉格朗日恆等式的證明
法1 directly
證明：
\[左邊=\sum_{i=1}^{n}a_{i}^{2}b_{i}^{2}+\sum_{1 \leq i< j \leq n}a_{i}^{2}b_{j}^{2}+\sum_{1 \leq i< j \leq n}a_{j}^{2}b_{i}^{2} \]
\[右邊=\sum_{i=1}^{n}a_{i}^{2}b_{i}^{2}+2\sum_{1 \leq i< j \leq n}a_{i}b_{j}a_{j}b_{i}+\sum_{1 \leq i< j \leq n}a_{i}^{2}b_{j}^{2}+\sum_{1 \leq i< j \leq n}a_{j}^{2}b_{i}^{2}-2\sum_{1 \leq i< j \leq n}a_{i}b_{j}a_{j}b_{i} \]
左邊=右邊
證畢#
法2 數學歸納法
證明：
\[1^{\circ}\quad當i=1時，左邊=a_{1}^{2}b_{1}^{2}=右邊，滿足原式 \]
\[2^{\circ}\quad假設當i=n-1時成立，則有(\sum_{i=1}^{n-1}a_{i}^{2})(\sum_{i=1}^{n-1}b_{i}^{2})=(\sum_{i=1}^{n-1}a_{i}b_{i})^{2}+\sum_{1 \leq i< j \leq n-1}(a_{i}b_{j}-a_{j}b_{i})^{2} \]
\[3^{\circ}\quad當i=n時，(\sum_{i=1}^{n}a_{i}^{2})(\sum_{i=1}^{n}b_{i}^{2})=(\sum_{i=1}^{n-1}a_{i}^{2})(\sum_{i=1}^{n-1}b_{i}^{2})+a_{n}^{2}+b_{n}^{2}+\sum_{i=1}^{n-1}a_{i}^{2}b_{n}^{2}+\sum_{i=1}^{n-1}b_{i}^{2}a_{n}^{2} \]
\[(\sum_{i=1}^{n}a_{i}b_{i})^{2}+\sum_{1 \leq i< j \leq n}(a_{i}b_{j}-a_{j}b_{i})^{2}=(\sum_{i=1}^{n-1}a_{i}b_{i})^{2}+2a_{n}b_{n}\sum_{i=1}^{n-1}a_{i}b_{i}+a_{n}^{2}+b_{n}^{2}+\sum_{1 \leq i< j \leq n-1}(a_{i}b_{j}-a_{j}b_{i})^{2}-2a_{n}b_{n}\sum_{i=1}^{n-1}a_{i}b_{i} \]
\[可得(\sum_{i=1}^{n}a_{i}^{2})(\sum_{i=1}^{n}b_{i}^{2})=(\sum_{i=1}^{n}a_{i}b_{i})^{2}+\sum_{1 \leq i< j \leq n}(a_{i}b_{j}-a_{j}b_{i})^{2} \]
證畢#