Question

观察下列式子 1/1×2=1-1/2 1/2×3=1/2-1/3 1/3×4

观察下列式子 1/1×2=1-1/2 1/2×3=1/2-1/3 1/3×4=1/3-1/4 ……
已知 |a-1|+|ab-3|=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+98)(b+98)
求解 在线等 急!要过程

Answer 1

绝对值项恒非负，两非负项之和=0，两非负项都=0
a-1=0
ab-3=0
解得a=1 b=3
b=a+2
1/(ab)+1/[(a+1)(b+1)]+1/[(a+2)(b+2)]+...+1/[(a+98)(b+98)]
=1/[a(a+2)]+1/[(a+1)(a+3)]+1/[(a+2)(a+4)]+...+1/[(a+98)(a+100)]
=(1/2)[1/a-1/(a+2)+1/(a+1)-1/(a+3)+1/(a+2)-1/(a+4)+...+1/(a+98)-1/(a+100)]
=(1/2)[1/a+1/(a+1)+...+1/(a+98)-(1/(a+2)+1/(a+3)+...+1/(a+100))]
=(1/2)[1/a+1/(a+1)-1/(a+99)-1/(a+100)]
=(1/2)(1/1+1/2-1/100-1/101)
=14949/202

Answer 2

|a-1|+|ab-3|=0,可得a=1；b=3

1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+98)(b+98)

可得到：
1/3+[1/(a+1)-1/(b+1)]/2+[1/(a+2)-1/(b+2)]/2+。。。。[1/(a+98)-1/(b+98)]/2
最后得到1/3+0.5[1/(a+1)+1/(a+2)+1/(a+98)]-0.5[1/(b+1)+1/(b+2)+1/(b+98)]
=1/3+0.5(1/2+1/3-1/100-1/101)=14949/20200
约等于0.74~