若二次函数y=1/2(x^2-100x+196+Ix^2-100x+196I）,则当自变量x取1.2.3.100这100个自然数时,函数的

Question

若二次函数y=1/2(x^2-100x+196+Ix^2-100x+196I）,则当自变量x取1.2.3.100这100个自然数时,函数的

Answer 1

当2≤x≤98时,因为 x^2-100x+196=(x-2)*(x-98)≤0,
所以恒有 y=[x^2-100x+196-(x^2-100x+196)]/2=0,
当x=1,99,100时,y=[x^2-100x+196+(x^2-100x+196)]/2=x^2-100x+196.
y(1)=y(99)=97,y(100)=196.
所以：y(1)+y(2)+y(3)+y(4)+……+y(97)+y(98)+y(99)+y(100)
=97+0+0+0+……+0+0+97+196=390.