Ciąg jest geometryczny, jeżeli [tex]q=\text{const}[/tex].
[tex]q=\dfrac{a_{n+1}}{a_n}\\\\\\a_n=3^{2n+1}-9^{n-1}\\\\a_{n+1}=3^{2(n+1)+1}-9^{n+1-1}=3^{2n+2+1}-9^n=3^{2n+3}-9^n\\\\\\q=\dfrac{3^{2n+3}-9^n}{3^{2n+1}-9^{n-1}}=\dfrac{3^{2n}\cdot3^3-3^{2n}}{3^{2n}\cdot 3-3^{2n}\cdot 9^{-1}}=\dfrac{3^{2n}(27-1)}{3^{2n}\left(3-\dfrac{1}{9}\right)}=\dfrac{26}{\dfrac{26}{9}}=9=\text{const}[/tex]
