Ambele siruri nu au primul termen 0, si verifica conditia: fiecare termen, incepand cu al doilea este medie geometrica a vecinilor sai.
1)[tex] a_{n-1}* a_{n+1} =2^{-n-1+1}* 2^{-n+1+1}= 2^{-2n+2} =(2^{-n+1})^2= a_{n}^2 [/tex]
2)[tex]a_{n-1}*a_{n+1} = \frac{2^{n-1} }{ 3^{n-1} }* \frac{ 2^{n+1} }{ 3^{n+1} }= \frac{ 2^{2n} }{ 3^{2n} }=( \frac{ 2^{n} }{3^n})^2= a_{n}^2 [/tex]
