\(\Sigma_{n=0}P_n=1P_0+\left(\frac{10}{4}\right)P_0+\left(\frac{10}{4}\right)^2P_0+\left(\frac{10}{4}\right)^3P_0+\left(\frac{10}{4}\right)^3\left(\frac{10}{8}\right)P_0\)\(\left(\frac{10}{4}\right)^3\left(\frac{10}{8}\right)^2P_0+\left(\frac{10}{4}\right)^3\left(\frac{10}{8}\right)^3P_0+\left(\frac{10}{4}\right)^3\left(\frac{10}{8}\right)^3\left(\frac{10}{12}\right)^{n-6}P_0=1\)
\(P_0=1+\left(\frac{10}{4}\right)+\left(\frac{10}{4}\right)^2+\left(\frac{10}{4}\right)^3+\left(\frac{10}{4}\right)^3\left(\frac{10}{8}\right)+\left(\frac{10}{4}\right)^3\left(\frac{10}{8}\right)^2+\left(\frac{10}{4}\right)^3\left(\frac{10}{8}\right)^3+\)\(\left(\frac{10}{4}\right)^3\left(\frac{10}{8}\right)^3\left(\frac{10}{12}\right)^{n-6}=1\)
\(P_0=\left[69.31+\left(\frac{10}{4}\right)^3\left(\frac{10}{8}\right)^3+\left[1+\left(\frac{10}{12}\right)+\left(\frac{10}{12}\right)^2\right]\right]\)
\(P_0\left[69.31+\left(\frac{10}{4}\right)^3\left(\frac{10}{8}\right)^3\left[\frac{1}{1-\left(\frac{10}{12}\right)}\right]\right]=1\)
\(P_0=\frac{1}{69.31+\left(\frac{10}{4}\right)^3\left(\frac{10}{8}\right)^3\left[\frac{1}{1-\left(\frac{10}{12}\right)}\right]}=3.9615\)

Resultado.

\(P_0=3.9615\)