1、有一个适合中学生的拉格朗日恒等式: (a1)2+(a2)2(b1)2+(b2)2= (a1)(b1)+(a2)(b2)2+(a2)(b1)-(a1)(b2)2 (a1)2+(a2)2+(a3)2(b1)2+(b2)2+(b3)2= =(a1)(b1)+(a2)(b2)+(a3)(b3)2+(a2)(b1)-(a1)(b2)2+ +(a3)(b1)-(a1)(b3)2+(a2)(b3)-(a3)(b2)2 (a1)2+.+(an)2(b1)2+.+(bn)2= =(a1)(b1)+.+(an)(bn)2+(a2)(b1)-(a1)(b2)2+ +(a3)(b1)-(a1)(b3)2+(a(n-1
2、)(bn)-(an)(b(n-1)2 用数学归纳法证明. 1. 显然 n=1时,(a1)2(b1)2=(a1)(b1)2. 拉格朗日恒等式成立. 2. 设 n=k时,拉格朗日恒等式成立. 当 n=k+1时, (a1)2+.+(a(n+1)2(b1)2+.+(b(n+1)2- -(a1)(b1)+.+(a(n+1)(b(n+1)2= =(a1)2+.+(an)2(b1)2+.+(bn)2- -(a1)(b1)+.+(an)(bn)2+ +(a(n+1)2(b1)2+(b(n+1)2(a1)2+ +(a(n+1)2(bn)2+(b(n+1)2(an)2- -2a(n+1)b(n+1)(a1)(b1
3、)+.+(an)(bn)= =(a2)(b1)-(a1)(b2)2+(a3)(b1)-(a1)(b3)2+ +(a(n-1)(bn)-(an)(b(n-1)2+ +(a(n+1)2(b1)2-2a(n+1)b(n+1)(a1)(b1)+ +(b(n+1)2(a1)2+(a(n+1)2(bn)2- -2a(n+1)b(n+1)(an)(bn)+(b(n+1)2(an)2= =(a2)(b1)-(a1)(b2)2+(a3)(b1)-(a1)(b3)2+ +(a(n-1)(bn)-(an)(b(n-1)2+ +(a(n+1)(b1)-b(n+1)(a1)2+ +(a(n+1)(bn)-b(n+1)(bn)2 所以 n=k+1时,拉格朗日恒等式成立. 这样数学归纳法证明了拉格朗日恒等式.
