解：设f(x)=ax^2+bx+c由f(0)=1，可得：c=1即f(x)=ax^2+bx+1f(x+1)=a(x+1)^2+b(x+1)+1=ax^2+(2a+b)x+(1+a+b)f(x+1)-f(x)=2ax+(a+b)因f(x+1)-f(x)=2x那么含x的多项式2ax+(a+b)与2x的系数相等则2a=2,a+b=0求出a=1,b=-1
