Question

已知二次函数fx满足f（x+1）-fx=2x且f0=1求fx?

Answer 1

f(x)=ax��+bx+c
f(0)=0+0+c=1
c=1
f(x+1)-f(x0
=a(x+1)��+b(x+1)+c-ax��-bx-c
=2ax+a+b
=2x
所以2a=2
a+b=0
所以a=1b=-1
所以f(x)=x��-x+1