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Question Number 81369 by mr W last updated on 12/Feb/20

Commented by mr W last updated on 12/Feb/20

$$ThisisarepostofQ81308.$$$$$$$$Ifoundtheanswerthroughthe$$$$followingway,butitseemslengthy$$$$andcomplex.Arethereanyother$$$$standardmethods?$$

Commented by mr W last updated on 12/Feb/20

Commented by MJS last updated on 12/Feb/20

$$\mathrm{I}\mathrm{think}\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{better}\mathrm{method},\mathrm{but}\mathrm{you}$$$$\mathrm{made}\mathrm{a}\mathrm{consequent}\mathrm{typo}134\mathrm{instead}\mathrm{of}143$$

Commented by MJS last updated on 12/Feb/20

$$\mathrm{can}\mathrm{you}\mathrm{find}\mathrm{the}\mathrm{last}n\mathrm{with}{x}_n=0?$$

Commented by mr W last updated on 12/Feb/20

$$thankyouforreviewing,andpointing$$$$outthetypo!letmethinkingaboutthe$$$$questionabout{x}_n=0.$$

Commented by MJS last updated on 12/Feb/20

$$\mathrm{I}\mathrm{think}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{at}n=108$$

Commented by mr W last updated on 12/Feb/20

$$yes!x_{108}isthelastzeroterm.$$

Answered by MJS last updated on 12/Feb/20

$$2104$$

Answered by M±th+et£s last updated on 12/Feb/20

$$usingZtransfrom$$$$x_0=x_2=...=x_{11}=0,x_{12}=2,x_{n+13}=x_{n+4}+2x_n$$$$Z^{13}x\left(z\right)-2z=z^{4}x\left(z\right)+2x\left(z\right)\to$$$$x\left(z\right)=\frac{2z}{z^{13}-z^4-2}=\frac{2z^{-12}}{1+z^{-9}-2z^{-13}}=\underset{k=1}{\sum ^{13}}\frac{A_k}{1-p_z{z}^{-1}}$$$$x_n=\underset{k=1}{\sum ^{13}}{A}_k{p}_k^n\to x_{142}=\underset{k=1}{\sum ^{142}}{A}_k{p}_k^n=2104$$

Commented by mr W last updated on 13/Feb/20

$$thankyou!$$$$ihavenotunderstoodyet.canyou$$$$pleasecheckifthereareanytypos!$$$$pleasegivemoreexplanationto$$$$yourlastlineandhowyougot2104.$$

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