设f(x)=ax²+bx+c f(x+1)+f(x-1) =2ax²+2bx+2a+2c
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![设f(x)=ax²+bx+c f(x+1)+f(x-1) =2ax²+2bx+2a+2c](/uploads/image/z/2832544-64-4.jpg?t=%E8%AE%BEf%28x%29%3Dax%26%23178%3B%2Bbx%2Bc+f%28x%2B1%29%2Bf%28x-1%29+%3D2ax%26%23178%3B%2B2bx%2B2a%2B2c)
设f(x)=ax²+bx+c f(x+1)+f(x-1) =2ax²+2bx+2a+2c
设f(x)=ax²+bx+c f(x+1)+f(x-1) =2ax²+2bx+2a+2c
设f(x)=ax²+bx+c f(x+1)+f(x-1) =2ax²+2bx+2a+2c
f(x)=ax²+bx+c
故:f(x+1)+f(x-1)
=a(x+1)²+b(x+1)+c +a(x-1)²+b(x-1)+c
=2ax²+2bx+2a+2c