Răspuns :
numere consecutive: 1 / n · ( n + 1) = 1 / n - 1 / ( n +1)
1 / 1 · 2 = 1 /1 - 1 / 2
1 / 2 ·3 = 1 / 2 - 1 / 3
1 / 3 · 4 = 1 / 3 - 1 / 4
.................................................
1 / 2005 · 2006 = 1 /2005 - 1 / 2006
-------------------------------------------------------
adunam = raman primul si ultimul termenu ,
= 1 / 1 - 1 /2006 = 2005 / 2006
1 / 1 · 2 = 1 /1 - 1 / 2
1 / 2 ·3 = 1 / 2 - 1 / 3
1 / 3 · 4 = 1 / 3 - 1 / 4
.................................................
1 / 2005 · 2006 = 1 /2005 - 1 / 2006
-------------------------------------------------------
adunam = raman primul si ultimul termenu ,
= 1 / 1 - 1 /2006 = 2005 / 2006
[tex]\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2005\cdot2006}=\\
1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2005}-\frac{1}{2006}=\\
1-\frac{1}{2006}=\frac{2005}{2006}[/tex]