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> 函数y=f（x）对任意实数x，y都有f（x+y）=f（x）+f（y）+2xy．（1）求f（0）的值；（2）若f（1）=1，求f（2），f（3），f（4）的值，猜想f（n）的表达式并用数学归纳法证明你的结论；（3）若f（1）≥1，求证： f( 1 2n )＞0(n∈N*)．

问题：

发布时间：2024-09-12

函数y=f（x）对任意实数x，y都有f（x+y）=f（x）+f（y）+2xy．
（1）求f（0）的值；
（2）若f（1）=1，求f（2），f（3），f（4）的值，猜想f（n）的表达式并用数学归纳法证明你的结论；
（3）若f（1）≥1，求证： f(

1

2n

)＞0(n∈N*)．

参考答案：

提问用户：nP***wuA

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函数y=f（x）对任意实数x，y都有f（x+y）=f（x）+f（y）+2xy． （1）求f（0）的值； （2）若f（1）=1，求f（2），f（3），f（4）的值，猜想f（n）的表达式并用数学归纳法证明你的结论； （3）若f（1）≥1，求证： f( 1 2n )＞0(n∈N*)．

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