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Question Number 184401 by mr W last updated on 06/Jan/23

U_n = ((((−4)^(n+1) −1)/(1−(−4)^n )))U_(n−1) with U_0 =1 find U_(n ) in terms of n (question Q173132 reposted)

$${U}_{{n}} \:=\:\left(\frac{\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} −\mathrm{1}}{\mathrm{1}−\left(−\mathrm{4}\right)^{{n}} }\right){U}_{{n}−\mathrm{1}} \:{with}\:{U}_{\mathrm{0}} =\mathrm{1} \\ $$$${find}\:{U}_{{n}\:} \:{in}\:{terms}\:{of}\:{n}\:\: \\ $$$$ \\ $$$$\left({question}\:{Q}\mathrm{173132}\:{reposted}\right) \\ $$

Commented by Frix last updated on 06/Jan/23

Sir have you tried Q178468?

$$\mathrm{Sir}\:\mathrm{have}\:\mathrm{you}\:\mathrm{tried}\:{Q}\mathrm{178468}? \\ $$

Commented by mr W last updated on 07/Jan/23

not yet sir.

$${not}\:{yet}\:{sir}. \\ $$

Commented by Frix last updated on 07/Jan/23

I tried but could not find any path...

$$\mathrm{I}\:\mathrm{tried}\:\mathrm{but}\:\mathrm{could}\:\mathrm{not}\:\mathrm{find}\:\mathrm{any}\:\mathrm{path}... \\ $$

Answered by witcher3 last updated on 06/Jan/23

(U_n /U_(n−1) )=−((1−(−4)^(n+1) )/(1−(−4)^n )),a_n =1−(−4)^n (U_k /U_(k−1) )=−(a_(k+1) /a_k ) Π_(k=1) ^n (U_k /U_(k−1) )=Π_(k=1) ^n −(a_(k+1) /a_k )⇔(U_n /U_0 )=(−1)^n .(a_(n+1) /a_1 )=(((−1)^n )/5)(1−(−4)^(n+1) ) ⇔u_n =(((−1)^n )/5)(1−(−4)^(n+1) )

$$\frac{{U}_{{n}} }{{U}_{{n}−\mathrm{1}} }=−\frac{\mathrm{1}−\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} }{\mathrm{1}−\left(−\mathrm{4}\right)^{\boldsymbol{{n}}} },{a}_{{n}} =\mathrm{1}−\left(−\mathrm{4}\right)^{{n}} \\ $$$$\frac{{U}_{{k}} }{{U}_{{k}−\mathrm{1}} }=−\frac{{a}_{{k}+\mathrm{1}} }{{a}_{{k}} } \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{{U}_{{k}} }{{U}_{{k}−\mathrm{1}} }=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}−\frac{{a}_{{k}+\mathrm{1}} }{{a}_{{k}} }\Leftrightarrow\frac{{U}_{{n}} }{{U}_{\mathrm{0}} }=\left(−\mathrm{1}\right)^{{n}} .\frac{{a}_{{n}+\mathrm{1}} }{{a}_{\mathrm{1}} }=\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{5}}\left(\mathrm{1}−\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} \right) \\ $$$$\Leftrightarrow{u}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{5}}\left(\mathrm{1}−\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} \right) \\ $$$$ \\ $$

Commented by witcher3 last updated on 06/Jan/23

y′re Welcom have a Nice Day

$${y}'{re}\:{Welcom}\:{have}\:{a}\:{Nice}\:{Day} \\ $$

Commented by mr W last updated on 06/Jan/23

thanks! great solution!

$${thanks}!\:{great}\:{solution}! \\ $$

Answered by mr W last updated on 06/Jan/23

U_n = ((((−4)^(n+1) −1)/(1−(−4)^n )))U_(n−1) (U_n /((−4)^(n+1) −1))=−(U_(n−1) /((−4)^n −1)) (U_(n−1) /((−4)^n −1))=−(U_(n−2) /((−4)^(n−1) −1)) ...... (U_1 /((−4)^2 −1))=−(U_0 /((−4)^1 −1)) (U_n /((−4)^(n+1) −1))=(−1)^n (U_0 /((−4)^1 −1))=(((−1)^n )/(−5)) ⇒U_n =(((−1)^n ((−4)^(n+1) −1))/(−5)) =((4^(n+1) +(−1)^n )/5) ✓

$${U}_{{n}} \:=\:\left(\frac{\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} −\mathrm{1}}{\mathrm{1}−\left(−\mathrm{4}\right)^{{n}} }\right){U}_{{n}−\mathrm{1}} \\ $$$$\frac{{U}_{{n}} }{\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} −\mathrm{1}}=−\cancel{\frac{{U}_{{n}−\mathrm{1}} }{\left(−\mathrm{4}\right)^{{n}} −\mathrm{1}}} \\ $$$$\cancel{\frac{{U}_{{n}−\mathrm{1}} }{\left(−\mathrm{4}\right)^{{n}} −\mathrm{1}}}=−\cancel{\frac{{U}_{{n}−\mathrm{2}} }{\left(−\mathrm{4}\right)^{{n}−\mathrm{1}} −\mathrm{1}}} \\ $$$$...... \\ $$$$\cancel{\frac{{U}_{\mathrm{1}} }{\left(−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{1}}}=−\frac{{U}_{\mathrm{0}} }{\left(−\mathrm{4}\right)^{\mathrm{1}} −\mathrm{1}} \\ $$$$\frac{{U}_{{n}} }{\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} −\mathrm{1}}=\left(−\mathrm{1}\right)^{{n}} \frac{{U}_{\mathrm{0}} }{\left(−\mathrm{4}\right)^{\mathrm{1}} −\mathrm{1}}=\frac{\left(−\mathrm{1}\right)^{{n}} }{−\mathrm{5}} \\ $$$$\Rightarrow{U}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}} \left(\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} −\mathrm{1}\right)}{−\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}^{{n}+\mathrm{1}} +\left(−\mathrm{1}\right)^{{n}} }{\mathrm{5}}\:\checkmark \\ $$

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