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)

kuai

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

Answer 2

太难了，高中的玩意。我听也没听