e cam de liceu
se "observa " ca termenul general poate fi scris;
3/((n*(n+3))= 3* (1/3)*((1/n-1/(n+3))
3/(2*5)= 3* (1/3)(1/2-1/5)
3/(5*8)=3*(1/3)*(1/5-1/8)
3/(8*11)=3*(1/3)*(1/8-1/11)
................................................
3/(2012*2015)=3*(1/3)* (1/2012-1/2015)
adunand tot
ne da
1*(1/2-1/2015)=(2015-2)/ (2*2015)=2013/4030
