Question Number 117243 by A8;15: last updated on 10/Oct/20
Commented by A8;15: last updated on 10/Oct/20
please help sir
Answered by Olaf last updated on 10/Oct/20
$$ \\ $$$${u}_{{n}} \:=\:{x}+\frac{{n}+\mathrm{1}}{{u}_{{n}+\mathrm{1}} } \\ $$$${u}_{\mathrm{0}} \:=\:{x}+\frac{\mathrm{1}}{{u}_{\mathrm{1}} }\:=\:{x}+\frac{\mathrm{1}}{{x}+\frac{\mathrm{2}}{{u}_{\mathrm{2}} }}\:=\:{x}+\frac{\mathrm{1}}{{x}+\frac{\mathrm{2}}{{x}+\frac{\mathrm{3}}{{u}_{\mathrm{3}} ...}}} \\ $$$$\mathrm{C}\:=\:{u}_{\mathrm{0}} \:=\:{x} \\ $$
Commented by A8;15: last updated on 10/Oct/20
the answer is C=x ?