Question

Determinati ratia q a unei progresii geometrice cu termeni strict pozitivi (bn)n>=1 care satisface conditia : b1+b3+...+b2n-1 supra b2+b4+...+b2n=3
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Answer 1

[tex]\it \dfrac{b_1+b_3+b_5+ ... +b_{2n-1}}{b_2+b_4+b_6+...+b_{2n}}=3 \Longrightarrow \dfrac{b_1(1+q^2+q^4+ ... + q^{2n-2})}{b_1(q+q^3+q^5+ ... +q^{2n-1})} =3 \\\;\\ \\\;\\ \Longrightarrow \dfrac{1+q^2+q^4+ ... + q^{2n-2}}{q(1+q^2+q^4+ ... + q^{2n-2} )} =3 \Longrightarrow \dfrac{1}{q} = 3 \Longrightarrow q= \dfrac{1}{3}[/tex]