抽象函数解析式解法一:令x=1,y=-1,代入:f(0)=f(1)=1;令y=1,代入原式:f(x+1)=f(x)+2(x+1)(需要注意的是x=0时,这个式子不成立)移项后:f(x+1)-f(x)=2x+2这个显然是个递推式,下面用n代替x进行演绎:显
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![抽象函数解析式解法一:令x=1,y=-1,代入:f(0)=f(1)=1;令y=1,代入原式:f(x+1)=f(x)+2(x+1)(需要注意的是x=0时,这个式子不成立)移项后:f(x+1)-f(x)=2x+2这个显然是个递推式,下面用n代替x进行演绎:显](/uploads/image/z/5365107-27-7.jpg?t=%E6%8A%BD%E8%B1%A1%E5%87%BD%E6%95%B0%E8%A7%A3%E6%9E%90%E5%BC%8F%E8%A7%A3%E6%B3%95%E4%B8%80%EF%BC%9A%E4%BB%A4x%3D1%2Cy%3D-1%2C%E4%BB%A3%E5%85%A5%EF%BC%9Af%280%29%3Df%281%29%3D1%3B%E4%BB%A4y%3D1%2C%E4%BB%A3%E5%85%A5%E5%8E%9F%E5%BC%8F%EF%BC%9Af%28x%2B1%29%3Df%28x%29%2B2%28x%2B1%29%EF%BC%88%E9%9C%80%E8%A6%81%E6%B3%A8%E6%84%8F%E7%9A%84%E6%98%AFx%3D0%E6%97%B6%2C%E8%BF%99%E4%B8%AA%E5%BC%8F%E5%AD%90%E4%B8%8D%E6%88%90%E7%AB%8B%EF%BC%89%E7%A7%BB%E9%A1%B9%E5%90%8E%EF%BC%9Af%28x%2B1%29-f%28x%29%3D2x%2B2%E8%BF%99%E4%B8%AA%E6%98%BE%E7%84%B6%E6%98%AF%E4%B8%AA%E9%80%92%E6%8E%A8%E5%BC%8F%2C%E4%B8%8B%E9%9D%A2%E7%94%A8n%E4%BB%A3%E6%9B%BFx%E8%BF%9B%E8%A1%8C%E6%BC%94%E7%BB%8E%EF%BC%9A%E6%98%BE)
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