Question Number 32995 by abdo imad last updated on 09/Apr/18
$${let}\:{u}_{\mathrm{0}} ={a}\:,\:{u}_{\mathrm{1}} ={b}\:{and}\:{u}_{{n}+\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left({u}_{{n}} \:+{u}_{{n}+\mathrm{1}} \right) \\ $$$$\left.\mathrm{1}\right)\:{find}\:{u}_{{n}} \:{interms}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow\infty} \:{u}_{{n}} \:\:{if}\:{a}=\mathrm{0} \\ $$$$ \\ $$
Commented by abdo imad last updated on 10/Apr/18
$${u}_{{n}+\mathrm{2}} \:=\frac{\mathrm{1}}{\mathrm{2}}\left({u}_{{n}} \:+{u}_{{n}+\mathrm{1}} \right)\:\Rightarrow\mathrm{2}{u}_{{n}+\mathrm{2}} \:={u}_{{n}\:\:} +{u}_{{n}+\mathrm{1}\:} \:\Rightarrow \\ $$$$\mathrm{2}{u}_{{n}+\mathrm{2}} \:−{u}_{{n}+\mathrm{1}} \:−{u}_{{n}} =\mathrm{0}\:\:{the}\:{caracteristic}\:{equation}\:{is} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \:−{x}−\mathrm{1}=\mathrm{0}\:{we}\:{have}\:\Delta=\mathrm{1}−\mathrm{4}\left(\mathrm{2}\right)\left(−\mathrm{1}\right)=\mathrm{9}\:\Rightarrow \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{3}}{\mathrm{4}}\:=\mathrm{1}\:{and}\:{x}_{\mathrm{2}} =\:\frac{\mathrm{1}−\mathrm{3}}{\mathrm{4}}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{u}_{{n}} =\alpha\:+\beta\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$${u}_{\mathrm{0}} ={a}\:\Rightarrow\alpha\:+\beta\:={a}\:,\:{u}_{\mathrm{1}} ={b}\:\Rightarrow\:\alpha\:−\frac{\beta}{\mathrm{2}}\:={b}\:\Rightarrow \\ $$$$\mathrm{3}\frac{\beta}{\mathrm{2}}\:={a}−{b}\:\Rightarrow\beta=\frac{\mathrm{2}}{\mathrm{3}}\left({a}−{b}\right)\:,\:\alpha\:={a}−\beta\:={a}\:−\frac{\mathrm{2}}{\mathrm{3}}\left({a}−{b}\right) \\ $$$$=\:\frac{{a}+\mathrm{2}{b}}{\mathrm{3}}\:\:\:\Rightarrow\:{u}_{{n}} =\frac{{a}\:+\mathrm{2}{b}}{\mathrm{3}}\:+\frac{\mathrm{2}{a}−\mathrm{2}{b}}{\mathrm{3}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$$\left.\mathrm{2}\right){a}=\mathrm{0}\:\Rightarrow\:{u}_{{n}} =\frac{\mathrm{2}{b}}{\mathrm{3}}\:−\frac{\mathrm{2}{b}}{\mathrm{3}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \:\:\:{but}\:{lim}_{{n}\rightarrow\infty} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} =\mathrm{0}\Rightarrow \\ $$$${lim}_{{n}\rightarrow\infty} \:\:{u}_{{n}} =\frac{\mathrm{2}{b}}{\mathrm{3}}\:. \\ $$