Question

Fie a0,a1,a2...,an termenii unei progresii geometrice strict crescatoare de ratie q.Stiind ca a0=1 si an+1=3an-2an-1 sa se determine ratia q si suma Sn=a0+a1+...+an a primilor n+1 termeni ai progresiei.

Answer 1

[tex]a_{n+1} = a_{n}*q=x*q$ , $ a_{n}=a_{0}* q^{n+1} $, a_{0}=1 a_{n}=q^{n+1} q^{n+2} = 3q^{n+1} -2q^{n} $ impartim la q^{n} Avem : q^{2} = 3q - 2 ,$ q^{2}-3q+2=0, de unde q=1 sau q = 2 ,$ cum$ progresia$e$strict$crescatoare, q=2 [/tex]