1 1 1 1
-------------------- = ----------- [ --------- - -------------- ]
n ( n +2) 2 n n+2
fiecare termen devine o scadere cu 1/2 in fata
dam factor x si 1/2
1005 x + 1/2 [ 1 + 1/1- 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +............ 1/2009 - 1/ 2011]=
= 1005 / 2011
din paranteza raman doar trei termeni , ceilalti sunt zero
1005x+ 1/2 [ 1- 1/2011]= 1005 /2011
1005x+ 1/2 ·2010/ 2011=1005 /2011
1005·x= 1005 /2011 - 2010 / 2 · 2011
1005·x=( 2·1005 - 2010 ) 2·2011
1005· x=0
x=0
