$1+a+a^2+a^3+...+a^n=\dfrac{1-a^{n+1}}{1-a}$
áp dụng vào
$lim \dfrac{1+3+3^2+...+3^n}{1+2+2^2+...+2^n}=lim \dfrac{1-3^{n+1}}{-2}.\dfrac{-1}{1-2^{n+1}}=lim \dfrac{1-3^{n+1}}{2(1-2^{n+1})}=lim \dfrac{\dfrac{1}{3^{n+1}}-1}{\dfrac{2}{3^{n+1}}-2(\dfrac{2}{3})^{n+1}}=... $
$1+a+a^2+a^3+...+a^n=\dfrac{1-a^{n+1}}{1-a}$
áp dụng vào
$lim \dfrac{1+3+3^2+...+3^n}{1+2+2^2+...+2^n}=lim \dfrac{1-3^{n+1}}{-2}.\dfrac{-1}{1-2^{n+1}}=lim \dfrac{1-3^{n+1}}{2(1-2^{n+1})}=lim \dfrac{\dfrac{1}{3^{n+1}}-1}{\dfrac{2}{3^{n+1}}-2(\dfrac{2}{3})^{n+1}}=... $