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Question Number 73057 by mathmax by abdo last updated on 05/Nov/19
$${let}\:{P}_{{n}} =\left({x}+\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} −{x}^{\mathrm{2}{n}+\mathrm{1}} −\mathrm{1} \\ $$$${prove}\:{that}\:{x}^{\mathrm{2}} \:+{x}\:{divide}\:\underset{{n}} {{P}} \\ $$
Answered by MJS last updated on 06/Nov/19
$${x}^{\mathrm{2}} +{x}=\mathrm{0}\:\Rightarrow\:{x}=−\mathrm{1}\vee{x}=\mathrm{0} \\ $$$${P}_{{n}} \left(\mathrm{0}\right)=\mathrm{1}^{\mathrm{2}{n}+\mathrm{1}} −\mathrm{0}^{\mathrm{2}{n}+\mathrm{1}} −\mathrm{1}=\mathrm{0} \\ $$$${P}_{{n}} \left(−\mathrm{1}\right)=\mathrm{0}^{\mathrm{2}{n}+\mathrm{1}} −\left(−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\forall{n}\in\mathbb{N}^{\bigstar} :\:\left({x}^{\mathrm{2}} +{x}\right)\mid{P}_{{n}} \\ $$