Question

∑(i-1)²

已知|ab+2|+(a+1)²=0,求下列式子:[1/(a-1)(b+1)]+i1/(a-2)(b+2)]+……+[1/(a-100)(b+100)

Answer 1

ab+2=0,a+1=0,所以a=-1,b=2
原式=1/-2*3+1/-3*4+……1/-101*102
=-(1/2*3+1/3*4+……1/101*102)
=-(1/2-1/102)=-25/51