Răspuns:
[tex]S_n=b_1+b_2+\cdots+b_n,\;n\geq 1[/tex].
Atunci [tex]b_1=S_1=3+12=15[/tex], Pentru [tex]n\geq 2[/tex], avem
[tex]b_n=S_n-S_{n-1}=3^n+12-3^{n-1}-12=3^n-3^{n-1}=3^{n-1}\cdot (3-1)=3^{n-1}\cdot 2[/tex]
In particular, [tex]b_2=6,b_3=18[/tex]. Evident, [tex]b_2/b_1=6/15=3/5[/tex] dar [tex]b_3/b_2=3[/tex], deci sirul [tex](b_n)_n[/tex], [tex]n\geq 1[/tex] nu e o progresie geometrica.
