函数f(x)=2^x(ax^2+bx+c)满足f(x+1)-f(x)=2^x*x^2(x∈R),求常数a、b、c的值

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函数f(x)=2^x(ax^2+bx+c)满足f(x+1)-f(x)=2^x*x^2(x∈R),求常数a、b、c的值
函数f(x)=2^x(ax^2+bx+c)满足f(x+1)-f(x)=2^x*x^2(x∈R),求常数a、b、c的值

函数f(x)=2^x(ax^2+bx+c)满足f(x+1)-f(x)=2^x*x^2(x∈R),求常数a、b、c的值
f(x+1)=[2^(x+1)][a(x+1)^2+b(x+1)+c]
=2^x[2ax^2+(4a+2b)x+2a+2b+2c]
f(x+1)-f(x)=2^x[ax^2+(4a+b)x+2a+2b+c]
∵f(x+1)-f(x)=2^x·x^2
∴2^x·x^2=2^x[ax^2+(4a+b)x+2a+2b+c]
即x^2=ax^2+(4a+b)x+2a+2b+c
比较系数得:a=1;4a+b=0;2a+2b+c=0;
解得:a=1,b=-4,c=6