Question

2次函数f(x)满足f(x+1)-f(x)=2x,f(0)=1

Answer 1

f(x)为2次函数,设 f(x)=ax^2+bx+c

因为f(0)=1,所以c=1

f(x+1)-f(x)=a*(x+1)^2+b*(x+1)-a*x^2-b*x

=2ax+a+b ----> =2x

所以a=1 ,a+b=0 ---> b=-1

即 f(x)=x^2-x+1