数学函数问题？
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f(x)是二次函数
f(x)=ax^2+bx+c
f(0)=0
c=0
f(x+1)=f(x)+x+1
a(x+1)^2+b(x+1) =ax^2+bx +x+1
ax^2+(2a+b)x +a+b =ax^2+(b+1)x +1
coef. of x
2a+b= b+1
a=1/2
coef. of constant
a+b = 1
1/2+b=1
b=1/2
f(x) =(1/2)x^2 +(1/2)x
f(10)=50+5=55
f(x)=ax^2+bx+c
f(0)=0
c=0
f(x+1)=f(x)+x+1
a(x+1)^2+b(x+1) =ax^2+bx +x+1
ax^2+(2a+b)x +a+b =ax^2+(b+1)x +1
coef. of x
2a+b= b+1
a=1/2
coef. of constant
a+b = 1
1/2+b=1
b=1/2
f(x) =(1/2)x^2 +(1/2)x
f(10)=50+5=55
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