If la-1l + lab-2l = 0. Find 1 / AB + 1 / (a + 1) (B + 1) +... + 1 / (a + 2004) (B + 2004)

If la-1l + lab-2l = 0. Find 1 / AB + 1 / (a + 1) (B + 1) +... + 1 / (a + 2004) (B + 2004)

∵la-1l+lab-2l=0
∴a-1=0 ab-2=0
∴a=1 b=2
The original formula = 1 / (1 × 2) + 1 / (2 × 3) + 1 / (2005 × 2006)
=1-1/2+1/2-1/3+1/2005-1/2006
=1-1/2006
=2005/2006

Given lab + 2L + La + 1L + 0, then the algebraic formula 1 / (A-1) (B + 1) + 1 / (A-2) (B + 2) + ^ + 1 / (a-2009) (B + 2009)

a=-1,b=2 ,1/（a-1)(b+1) +1/（a-2)(b+2)+^+1/(a-2009)(b+2009)=-(1/2-1/3)-(1/3-1/4)--^--(1/3000-1/3001)=--（1/2--1/3+1/3--1/4+^+1/3000-1/3001)=--(1/2--1/3001)=--2009/6002

Given lab + 2L + La + 1L = 0, find 1 / (A-1) (B + 1) + 1 / (A-2) (B + 2)... + 1 / (2-2004) (B + 2004)

AB + 2 = 0, a + 1 = 0, so a = - 1, B = 2, so the original formula is equal to - 1 / (2x3) - 1 / (3x4) - 1 / (4x5)... - 1 / (2005x2006) minus sign. If we put forward it, we can get the split term in the bracket - [1 / (2x3) + 1 / (3x4) + 1 / (4x5)... + 1 / (2005x2006)], that is, 1 / (2x3) = 1 / 2-1 / 31 / (3x4) = 1 / 3-1 / 4