Question Number 159881 by LEKOUMA last updated on 22/Nov/21
$${Resolve}\: \\ $$$$\mathrm{1}.\:\:{u}_{{n}} −\mathrm{3}{u}_{{n}−\mathrm{1}} =\mathrm{12}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} \\ $$$$\mathrm{2}.\:{u}_{{n}} =\mathrm{2}{u}_{{n}−\mathrm{1}} +\mathrm{5cos}\:\left(\frac{{n}\Pi}{\mathrm{3}}\right),\:{u}_{{o}} =\mathrm{1}\: \\ $$$$\mathrm{3}.\:{u}_{{n}} ={u}_{{n}−\mathrm{1}} −{u}_{{n}−\mathrm{2}} +\mathrm{2sin}\:\left(\frac{{n}\Pi}{\mathrm{3}}\right) \\ $$$${with}\:{u}_{{o}} =\mathrm{1},\:{u}_{\mathrm{1}} =\mathrm{2} \\ $$