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

$$U_n=\left(\frac{{(-4)}^{n+1}-1}{1-{(-4)}^n}\right)U_{n-1}with{U}_0=1$$$$find{U}_{n}intermsofn$$$$$$$$\left(questionQ173132reposted\right)$$

Commented by Frix last updated on 06/Jan/23

$$\mathrm{Sir}\mathrm{have}\mathrm{you}\mathrm{tried}Q178468?$$

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

$$notyetsir.$$

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

$$\frac{U_n}{U_{n-1}}=-\frac{1-{(-4)}^{n+1}}{1-{(-4)}^{\mathit{n}}},a_n=1-{(-4)}^n$$$$\frac{U_k}{U_{k-1}}=-\frac{a_{k+1}}{a_k}$$$$\underset{k=1}{\prod ^n}\frac{U_k}{U_{k-1}}=\underset{k=1}{\prod ^n}-\frac{a_{k+1}}{a_k}\iff \frac{U_n}{U_0}={(-1)}^n.\frac{a_{n+1}}{a_1}=\frac{{(-1)}^n}{5}(1-{(-4)}^{n+1})$$$$\iff u_n=\frac{{(-1)}^n}{5}(1-{(-4)}^{n+1})$$$$$$

Commented by witcher3 last updated on 06/Jan/23

$$y^{\prime }reWelcomhaveaNiceDay$$

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

$$thanks!greatsolution!$$

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

$$U_n=\left(\frac{{(-4)}^{n+1}-1}{1-{(-4)}^n}\right)U_{n-1}$$$$\frac{U_n}{{(-4)}^{n+1}-1}=-\cancel{\frac{U_{n-1}}{{(-4)}^n-1}}$$$$\cancel{\frac{U_{n-1}}{{(-4)}^n-1}}=-\cancel{\frac{U_{n-2}}{{(-4)}^{n-1}-1}}$$$$......$$$$\cancel{\frac{U_1}{{(-4)}^2-1}}=-\frac{U_0}{{(-4)}^1-1}$$$$\frac{U_n}{{(-4)}^{n+1}-1}={(-1)}^n\frac{U_0}{{(-4)}^1-1}=\frac{{(-1)}^n}{-5}$$$$\Rightarrow U_n=\frac{{(-1)}^n({(-4)}^{n+1}-1)}{-5}$$$$=\frac{4^{n+1}+{(-1)}^n}{5}✓$$

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