Question Number 144300 by mathdanisur last updated on 24/Jun/21
Answered by Rasheed.Sindhi last updated on 25/Jun/21
$$\left({x}+\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} −{a}^{\mathrm{2}{n}+\mathrm{1}} ={x}\left({x}+\mathrm{1}\right){Q}\left({x}\right)+{ux}+{v} \\ $$$${For}\:{x}\in\mathbb{Z} \\ $$$${For}\:{x}=\mathrm{0} \\ $$$$\mathrm{1}−{a}^{\mathrm{2}{n}+\mathrm{1}} ={v} \\ $$$${For}\:{x}=−\mathrm{1} \\ $$$$−{a}^{\mathrm{2}{n}+\mathrm{1}} =−{u}+{v} \\ $$$$−{a}^{\mathrm{2}{n}+\mathrm{1}} =−{u}+\mathrm{1}−{a}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$−{u}+\mathrm{1}=\mathrm{0}\Rightarrow{u}=\mathrm{1} \\ $$$${R}\left({x}\right)={ux}+{v}=\left(\mathrm{1}\right){x}+\mathrm{1}−{a}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:={x}+\mathrm{1}−{a}^{\mathrm{2}{n}+\mathrm{1}} \\ $$
Commented by mathdanisur last updated on 25/Jun/21
$${thanks}\:{Sir}\:{cool} \\ $$