Question Number 31032 by abdo imad last updated on 02/Mar/18
$${let}\:{give}\:{x}_{\mathrm{0}} =\mathrm{0}\:,{y}_{\mathrm{0}} =\mathrm{1}\:{and}\:\left\{_{{y}_{{n}} ={x}_{{n}−\mathrm{1}} \:+{y}_{{n}−\mathrm{1}} } ^{{x}_{{n}} ={x}_{{n}−\mathrm{1}} \:−{y}_{{n}−\mathrm{1}} } \:\:\:\:\:\:{for}\:{n}\geqslant\mathrm{1}\:{let}\right. \\ $$$${z}_{{n}} ={x}_{{n}} \:+{i}\:{y}_{{n}} \:\:\:\:\:\forall{n}\in{N} \\ $$$$\left.\mathrm{1}\right){calculate}\:{z}_{\mathrm{0}} \:,{z}_{\mathrm{1}} \:{and}\:{z}_{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:\forall{n}\in{N}^{} ,{n}\geqslant\mathrm{1}\:\:{z}_{{n}} =\left(\mathrm{1}+{i}\right){z}_{{n}−\mathrm{1}} \:{find}\:{z}_{{n}} {then} \\ $$$${find}\:{the}\:{expression}\:{of}\:{x}_{{n}} \:{and}\:{y}_{{n}} \\ $$$$\left.\mathrm{3}\right){let}\:{put}\:{S}_{{n}} ={z}_{\mathrm{0}} \:+{z}_{\mathrm{1}} \:+….{z}_{{n}} \\ $$$${s}_{{n}} ={x}_{\mathrm{0}} \:+{x}_{\mathrm{1}} \:+…+{x}_{{n}} \\ $$$${s}_{{n}} ^{'} ={y}_{\mathrm{0}} \:+{y}_{\mathrm{1}} \:+…+{y}_{{n}} \:\:{find}\:{those}\:{sum}\:{interms}\:{of}\:{n}. \\ $$$$ \\ $$