Fie n si k nr. naturale
S=1+n+n^2+n^3+...+n^k
Inmultim cu n:
nxS=n+n^2+n^3+...+n^k+n(k+1)
nxS=(1+n+n^2+n^3+...+n^k)+n(k+1) -1
nxS=S+n(k+1) -1
nxS-S=n(k+1) -1
S(n-1)=n(k+1) -1
S=(n(k+1) -1)/(n-1)
a)
n=3 si k=99
1+3+3^2+...+3^99=(3^100 -1)/2
2*(1+3+3^2+...+3^99)=2*(3^100 -1)/2=
3^100 -1<3^100
b)
n=4, k=2013
1+4+4^2+...+4^2013=(4^2014 -1)/3
3*(1+4+4^2+...+4^2013)=3*(4^2014 -1)/3=
4^2014 -1<4^2014
