Question Number 4638 by Yozzii last updated on 17/Feb/16
$${Let}\:\boldsymbol{\mathrm{P}}=\begin{pmatrix}{\mathrm{1}−{p}_{\mathrm{1}} }&{{p}_{\mathrm{2}} }\\{{p}_{\mathrm{1}} }&{\mathrm{1}−{p}_{\mathrm{2}} }\end{pmatrix}=\begin{pmatrix}{{a}_{\mathrm{1},\mathrm{1}} }&{{a}_{\mathrm{1},\mathrm{2}} }\\{{a}_{\mathrm{2},\mathrm{1}} }&{{a}_{\mathrm{2},\mathrm{2}} }\end{pmatrix} \\ $$$${and}\:{that}\:\boldsymbol{\mathrm{P}}^{{n}} \:{be}\:{the}\:{nth}\:{power}\:{of}\:\boldsymbol{\mathrm{P}}\:{evaluated} \\ $$$${as}\:\boldsymbol{\mathrm{P}}^{{n}} =\boldsymbol{\mathrm{P}}×\boldsymbol{\mathrm{P}}^{{n}−\mathrm{1}} ;{i}.{e}\:{by}\:{successive}\: \\ $$$${pre}−{multiplication}\:{of}\:\boldsymbol{\mathrm{P}}\:{to}\:\boldsymbol{\mathrm{P}}^{\mathrm{2}} ,\boldsymbol{\mathrm{P}}^{\mathrm{3}} ,\boldsymbol{\mathrm{P}}^{\mathrm{4}} ,… \\ $$$${up}\:{to}\:\boldsymbol{\mathrm{P}}^{{n}−\mathrm{1}} .\:{Show}\:{that}\:{the}\:{element}\:{of} \\ $$$$\boldsymbol{\mathrm{P}}^{{n}} \:{in}\:{the}\:{second}\:{row}\:{and}\:{first}\:{column} \\ $$$${is}\:{a}_{\mathrm{2},\mathrm{1}} =\frac{{p}_{\mathrm{1}} \left(\mathrm{1}−\left(\mathrm{1}−{p}_{\mathrm{1}} −{p}_{\mathrm{2}} \right)^{{n}} \right)}{{p}_{\mathrm{1}} +{p}_{\mathrm{2}} }.\: \\ $$$$\left\{{The}\:{columns}\:{of}\:\boldsymbol{\mathrm{P}}^{{n}} \:{represent}\:\right. \\ $$$${probability}\:{vectors}\:{for}\:{all}\:{n}\in\mathbb{N}.\:{Hence}, \\ $$$$\left.{a}_{\mathrm{1},\mathrm{1}} +{a}_{\mathrm{2},\mathrm{1}} =\mathrm{1}\:{and}\:{a}_{\mathrm{1},\mathrm{2}} +{a}_{\mathrm{2},\mathrm{2}} =\mathrm{1}\:{for}\:{example}.\right\} \\ $$$$ \\ $$