Question Number 73537 by Rio Michael last updated on 13/Nov/19
$${prove}\:{by}\:{induction}\:{that}\:\mathrm{4}^{{n}} \:+\:\mathrm{3}^{{n}} \:+\mathrm{2}\:{is}\:{a}\:{multiple}\:{of}\:\mathrm{3} \\ $$$$\forall\:{n}\:{Z}^{+} \\ $$
Answered by mind is power last updated on 13/Nov/19
$${n}=\mathrm{1}\:{true} \\ $$$$\mathrm{4}+\mathrm{3}+\mathrm{2}=\mathrm{9} \\ $$$${suppose}\:\forall{n}\in\mathbb{N}−\left\{\mathrm{0}\right\}\:{u}_{{n}} =\mathrm{4}^{{n}} +\mathrm{3}^{{n}} +\mathrm{2}\:{is}\:{multiple}\:{of}\:\mathrm{3} \\ $$$${show}\:\mathrm{3}\mid{u}_{{n}+\mathrm{1}} \\ $$$${u}_{{n}+\mathrm{1}} =\mathrm{4}.\mathrm{4}^{{n}} +\mathrm{3}^{{n}} \ast\mathrm{3}+\mathrm{2} \\ $$$$=\mathrm{3}.\mathrm{4}^{{n}} +\mathrm{4}^{{n}} +\mathrm{3}^{{n}} +\mathrm{2}+\mathrm{2}.\mathrm{3}^{{n}} \\ $$$$=\mathrm{4}^{{n}} +\mathrm{3}^{{n}} +\mathrm{2}+\mathrm{3}.\mathrm{4}^{{n}} +\mathrm{3}^{{n}} .\mathrm{2} \\ $$$$={u}_{{n}} +\mathrm{3}\left(\mathrm{4}^{{n}} +\mathrm{2}.\mathrm{3}^{{n}−\mathrm{1}} \right) \\ $$$${since}\:{n}\geqslant\mathrm{1}\Rightarrow\mathrm{3}^{{n}−\mathrm{1}} \in\mathbb{N}\Rightarrow\mathrm{3}\mid\mathrm{3}.\left(\mathrm{4}^{{n}} +\mathrm{2}.\mathrm{3}^{{n}−\mathrm{1}} \right)\:{by}\:{hypothese}\:\mathrm{3}\mid{U}_{{n}} \Rightarrow \\ $$$$\mathrm{3}\mid{u}_{{n}+\mathrm{1}} \\ $$
Commented by Rio Michael last updated on 13/Nov/19
$${thank}\:{you}\:{sir} \\ $$