Question

已知|ab-2|与(b-1)2互为相反数，试求式子

1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005(b+2005)的值。

Answer 1

解 Iab-2|与(b-1)2互为相反数
所以Iab-2I=-(b-1)^2
b-1=0 b=1 ab-2=0 a=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005(b+2005)
=1/1*2+1/2*3+1/3*4+...+1/2006*2007
=1-1/2+1/2-1/3+1/3-1/4+...+1/2006-1/2007
=1-1/2007
=2006/2007

Answer 2

由|ab-2|与(b-1)2知道
b = 1
a = 2
b =a - 1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005(b+2005)
= 1/a(a-1) + 1/(a+1)a + 1/(a+1)(a+2)+...+1/(a+2004)(a+2005)
= 1/(a-1) - 1/a + 1/a - 1/(a+1)+...+1/(a+2004) - 1/(a+2005)
=1/(a-1) - 1/(a+2005)
= 1 - 1/2007
= 2006/2007