Question Number 37930 by zesearho last updated on 19/Jun/18
$${n}\in\mathbb{N} \\ $$$${U}_{{n}+\mathrm{1}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}+\mathrm{1}} +{U}_{{n}} \\ $$$${U}_{{n}} =? \\ $$
Commented by abdo.msup.com last updated on 19/Jun/18
$${we}\:{have}\:{u}_{{n}+\mathrm{1}} \:−{u}_{{n}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}+\mathrm{1}} \:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({u}_{{k}+\mathrm{1}} \:−{u}_{{k}} \right)=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{k}+\mathrm{1}} \Rightarrow \\ $$$${u}_{{n}} \:−{u}_{\mathrm{0}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{k}} \:=\frac{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}+\mathrm{1}} }{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\:−\mathrm{1} \\ $$$$=\mathrm{2}\:−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}+\mathrm{1}} \:−\mathrm{1}=\:\mathrm{1}−\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\Rightarrow \\ $$$${u}_{{n}} =\:{u}_{\mathrm{0}} \:+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:. \\ $$
Answered by ajfour last updated on 19/Jun/18
$${U}_{\mathrm{1}} =\:{U}_{\mathrm{0}} +\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${U}_{\mathrm{2}} ={U}_{\mathrm{1}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${U}_{{n}} =\:{U}_{\mathrm{0}} +\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} . \\ $$