Question

Ajutor la 5b

Va roooooooog

Answer 1

Răspuns:

$S_n=b_1+b_2+\cdots+b_n,\;n\geq 1$.

Atunci $b_1=S_1=3+12=15$, Pentru $n\geq 2$, avem

$b_n=S_n-S_{n-1}=3^n+12-3^{n-1}-12=3^n-3^{n-1}=3^{n-1}\cdot (3-1)=3^{n-1}\cdot 2$

In particular, $b_2=6,b_3=18$. Evident, $b_2/b_1=6/15=3/5$ dar $b_3/b_2=3$, deci sirul $(b_n)_n$, $n\geq 1$ nu e o progresie geometrica.