Dupa cum se observa, cele 2 sume au fiecare cate un numar cu acelasi numitor, formand perechi:
[tex]a+b=( \frac{1}{2} + \frac{1}{2} )+( \frac{1}{3} + \frac{2}{3} )+( \frac{1}{4} + \frac{3}{4} )+...+ (\frac{1}{2011} + \frac{2010}{2011} )\\
a+b= \frac{1+1}{2} + \frac{1+2}{3} + \frac{1+3}{4} +...+ \frac{1+2010}{2011}\\
a+b= \frac{2}{2} + \frac{3}{3} + \frac{4}{4} +... +\frac{2011}{2011}\\
a+b=1*2010=2010 \\
\frac{a+b}{2}= \frac{2010}{2}=1005 [/tex]
