求证拉格朗日恒等式

拉格朗日恒等式:

[(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))(bn)-(an)(b(n-1))]^2

证明：可以用数学归纳法
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)+...+(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