Question

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

Answer 1

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