已知函数f(x)是二次函数，且f(0)=0,f(x+1)=f(x)+x+1,求f(x)的解析式

设f(x)=ax^2+bx+c
f(0)=c=0
所以：f(x)=ax^2+bx
f(x+1)=f(x)+x+1
a(x+1)^2+b(x+1)=ax^2+bx+x+1
(2a-1)x+(a+b-1)=0
2a-1=0, a=1/2
a+b-1=0, b=1-a=1/2

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

f(x+1)-f(x)=x+1
f(x)-f(x-1)=x
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f(1)-f(0)=1
列加，f(x+1)=x(x+1)/2
f(x)=（x-1)x/2

设f(x)=ax^2+bx+c,
f(0)=0+0+c=0, c=0
f(0+1)=f(0)+0+1=1, f(1)=1, a+b=1,
f(0)=f(-1+1)=f(-1)-1+1=f(-1)=0, a-b=0, a=b=0.5
f(x)=0.5x^2+0.5x