f(0)= 0+ 1 =1
f(1)= 1+ 1 =2
....................................
f(1998)=1998+1 = 1999
adunind membru cu membru s= [f(0)+f(1)+.....+f(1998]=1+2+3+4+...+1999
S=1999(1999+1)/2 (suma primelor n numere naturale [tex] s_{n} = \frac{n(n+1)}{2} [/tex] )
N=2000 + 2 [tex] \frac{1999*2000}{2} [/tex]=2000+1999*2000=2000²
