Question

已知二次函数f(x)满足f(0)=0,且对任意x∈R总有f(x+1)=f(x)+x+1,求f(x)

Answer 1

f(x)=ax^2+bx+c,

0=f(0)=c.

f(x)=ax^2+bx,

1+x+f(x)=1+x+ax^2+bx=f(x+1)=a(x+1)^2+b(x+1)=ax^2+2ax+a+bx+b,

ax^2 + x(b+1) + 1 = ax^2 + x(b+2a) + a+b,

b+1=b+2a,a= 1/2.

1=a+b=1/2 + b,b=1/2.

f(x)=x(x+1)/2