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Question Number 26400 by abdo imad last updated on 25/Dec/17
$${find}\:{the}\:{sequence}\:\left({u}_{{n}} \right)\:{wich}\:{verify}\:{u}_{{n}} \:−\mathrm{2}\:{u}_{{n}−\mathrm{1}} \:+\mathrm{1}=\:\mathrm{2}^{{n}} \\ $$
Answered by prakash jain last updated on 25/Dec/17
$${x}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{2} \\ $$$${homogenous}\:{solution}={c}\mathrm{2}^{{n}} \\ $$$${particular}\:{solution} \\ $$$${u}_{{n}} −\mathrm{2}{u}_{{n}−\mathrm{1}} =\mathrm{2}^{{n}} −\mathrm{1} \\ $$$$\mathrm{Here}\:\mathrm{we}\:\mathrm{will}\:\mathrm{solve}\:\mathrm{for}\:\mathrm{2}\:\mathrm{parts} \\ $$$$\mathrm{part}\:\mathrm{I} \\ $$$${u}_{{n}} −\mathrm{2}{u}_{{n}−\mathrm{1}} =−\mathrm{1}\Rightarrow{u}_{{n}} \:\left(\mathrm{part}\:\mathrm{1}\right)=\mathrm{1} \\ $$$$\mathrm{part}\:\mathrm{II} \\ $$$${u}_{{n}} −\mathrm{2}{u}_{{n}−\mathrm{1}} =\mathrm{2}^{{n}} \\ $$$${u}_{{n}} ={an}\mathrm{2}^{{n}} \\ $$$${an}\mathrm{2}^{{n}} −{a}\left({n}−\mathrm{1}\right)\mathrm{2}^{{n}} =\mathrm{2}^{{n}} \\ $$$$\Rightarrow{a}=\mathrm{1} \\ $$$${particular}\:{solution} \\ $$$${u}_{{n}} ={n}\mathrm{2}^{{n}} +\mathrm{1} \\ $$$${u}_{{n}} ={c}\mathrm{2}^{{n}} +{n}\mathrm{2}^{{n}} +\mathrm{1} \\ $$