设x1x2……xn都是实数
证明:用归纳法来证明
(1)当n=2时
2(x1²+x2²)=(x1+x2)²
有:2x1²+2x2²=x1²+2x1x2+x2²
x1²-2x1x2+x2²=0
(x1-x2)²=0
那么x1=x2
(2)当n=3时
3(x1²+x2²+x3²)=(x1+x2+x3)²
有:3x1²+3x2²+3x3²=x1²+2x1x2+x2²+2x1x3+2x2x3+x3²
x1²-2x1x2...全部
证明:用归纳法来证明
(1)当n=2时
2(x1²+x2²)=(x1+x2)²
有:2x1²+2x2²=x1²+2x1x2+x2²
x1²-2x1x2+x2²=0
(x1-x2)²=0
那么x1=x2
(2)当n=3时
3(x1²+x2²+x3²)=(x1+x2+x3)²
有:3x1²+3x2²+3x3²=x1²+2x1x2+x2²+2x1x3+2x2x3+x3²
x1²-2x1x2+x2²+x1²-2x1x3+x3^2+x2²-2x2x3+x3²=0
(x1-x2)²+(x1-x3)²+(x2-x3)²=0
那么x1=x2=x3
(3)假设n(x1²+x2²+……+xn²)=(x1+x2+……+xn)² 时x1=x2=……=xn成立
(n+1)(x1²+x2²+……+xn²+x(n+1)²)=(x1+x2+……+xn+x(n+1))²
有n(x1+²x2²+……+xn²)+(x1²+x2²+……+xn²)+(n+1)[x(n+1)]²=(x1+x2+……+xn)² +2(x1+x2+……+xn)x(n+1)+[x(n+1)]²
即:(x1-x2)²+(x1-x2)²+……+(xn-x(n+1))²=0
有:x1=x2=……=xn=x(n+1)
证毕
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