取特殊值法设f(x)=ax^2+bx+1 令x=0 则f(1)=f(0)+2*0=1 a+b=0 令x=1 则f(2)=2+f(1)=34a+2b+1=3 解得a=1 b=-1 f(x)=x^2-x+1对照系数法设f(x)=ax^2+bx+1 f(x+1)=ax^2+2ax+bx+a+b+1f(x+1)=f(x)+2x=ax^2+(b+2)x+1对照系数...