Question Number 184401 by mr W last updated on 06/Jan/23
$${U}_{{n}} \:=\:\left(\frac{\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} −\mathrm{1}}{\mathrm{1}−\left(−\mathrm{4}\right)^{{n}} }\right){U}_{{n}−\mathrm{1}} \:{with}\:{U}_{\mathrm{0}} =\mathrm{1} \\ $$$${find}\:{U}_{{n}\:} \:{in}\:{terms}\:{of}\:{n}\:\: \\ $$$$ \\ $$$$\left({question}\:{Q}\mathrm{173132}\:{reposted}\right) \\ $$
Commented by Frix last updated on 06/Jan/23
$$\mathrm{Sir}\:\mathrm{have}\:\mathrm{you}\:\mathrm{tried}\:{Q}\mathrm{178468}? \\ $$
Commented by mr W last updated on 07/Jan/23
$${not}\:{yet}\:{sir}. \\ $$
Commented by Frix last updated on 07/Jan/23
$$\mathrm{I}\:\mathrm{tried}\:\mathrm{but}\:\mathrm{could}\:\mathrm{not}\:\mathrm{find}\:\mathrm{any}\:\mathrm{path}… \\ $$
Answered by witcher3 last updated on 06/Jan/23
$$\frac{{U}_{{n}} }{{U}_{{n}−\mathrm{1}} }=−\frac{\mathrm{1}−\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} }{\mathrm{1}−\left(−\mathrm{4}\right)^{\boldsymbol{{n}}} },{a}_{{n}} =\mathrm{1}−\left(−\mathrm{4}\right)^{{n}} \\ $$$$\frac{{U}_{{k}} }{{U}_{{k}−\mathrm{1}} }=−\frac{{a}_{{k}+\mathrm{1}} }{{a}_{{k}} } \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{{U}_{{k}} }{{U}_{{k}−\mathrm{1}} }=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}−\frac{{a}_{{k}+\mathrm{1}} }{{a}_{{k}} }\Leftrightarrow\frac{{U}_{{n}} }{{U}_{\mathrm{0}} }=\left(−\mathrm{1}\right)^{{n}} .\frac{{a}_{{n}+\mathrm{1}} }{{a}_{\mathrm{1}} }=\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{5}}\left(\mathrm{1}−\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} \right) \\ $$$$\Leftrightarrow{u}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{5}}\left(\mathrm{1}−\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} \right) \\ $$$$ \\ $$
Commented by witcher3 last updated on 06/Jan/23
$${y}'{re}\:{Welcom}\:{have}\:{a}\:{Nice}\:{Day} \\ $$
Commented by mr W last updated on 06/Jan/23
$${thanks}!\:{great}\:{solution}! \\ $$
Answered by mr W last updated on 06/Jan/23
$${U}_{{n}} \:=\:\left(\frac{\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} −\mathrm{1}}{\mathrm{1}−\left(−\mathrm{4}\right)^{{n}} }\right){U}_{{n}−\mathrm{1}} \\ $$$$\frac{{U}_{{n}} }{\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} −\mathrm{1}}=−\cancel{\frac{{U}_{{n}−\mathrm{1}} }{\left(−\mathrm{4}\right)^{{n}} −\mathrm{1}}} \\ $$$$\cancel{\frac{{U}_{{n}−\mathrm{1}} }{\left(−\mathrm{4}\right)^{{n}} −\mathrm{1}}}=−\cancel{\frac{{U}_{{n}−\mathrm{2}} }{\left(−\mathrm{4}\right)^{{n}−\mathrm{1}} −\mathrm{1}}} \\ $$$$…… \\ $$$$\cancel{\frac{{U}_{\mathrm{1}} }{\left(−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{1}}}=−\frac{{U}_{\mathrm{0}} }{\left(−\mathrm{4}\right)^{\mathrm{1}} −\mathrm{1}} \\ $$$$\frac{{U}_{{n}} }{\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} −\mathrm{1}}=\left(−\mathrm{1}\right)^{{n}} \frac{{U}_{\mathrm{0}} }{\left(−\mathrm{4}\right)^{\mathrm{1}} −\mathrm{1}}=\frac{\left(−\mathrm{1}\right)^{{n}} }{−\mathrm{5}} \\ $$$$\Rightarrow{U}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}} \left(\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} −\mathrm{1}\right)}{−\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}^{{n}+\mathrm{1}} +\left(−\mathrm{1}\right)^{{n}} }{\mathrm{5}}\:\checkmark \\ $$