Question Number 192425 by mehdee42 last updated on 17/May/23
$${Question} \\ $$$${if}\:\:“{k}''\:{is}\:{odd}\:\:\&\:{A}=\mathrm{1}^{{k}} +\mathrm{2}^{{k}} +…+{n}^{{k}\:\:} \:\&\:\:{B}=\mathrm{1}+\mathrm{2}+…+{n} \\ $$$${prove}\:{that}\:\::\:\:{B}\:\mid\:{A}\: \\ $$
Answered by MM42 last updated on 24/May/23
$${We}\:{khnow}\:\:;\:\:\mathrm{1}+\mathrm{2}+…+{n}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:\:\:\&\:\:\:\left({n},\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)=\mathrm{1}\:{or}\:\left(\frac{{n}}{\mathrm{2}},{n}+\mathrm{1}\right)=\mathrm{1} \\ $$$${Thus}\:,{it}\:{soffices}\:\:{to}\:\:{show}\:{that}\: \\ $$$${n}\mid{A}\:\&\:\frac{{n}+\mathrm{1}}{\mathrm{2}}\:\mid{A}\:\:{or}\:\:\frac{{n}}{\mathrm{2}}\mid{A}\:\&\:{n}+\mathrm{1}\mid{A} \\ $$$${proof}\:\therefore \\ $$$${if}\:\:“{n}''\:{is}\:{even}\:\Rightarrow \\ $$$$\:\frac{{n}}{\mathrm{2}}\mid\mathrm{1}^{{k}} +\left({n}−\mathrm{1}\right)^{{k}} \:,\:\mathrm{2}^{{k}} +\left({n}−\mathrm{2}\right)^{{k}} \:,\:…\:\left({i}\right) \\ $$$$\:{n}+\mathrm{1}\mid\mathrm{1}^{{k}} +{n}^{{k}} \:,\:\mathrm{2}^{{k}} +\left({n}−\mathrm{1}\right)^{{k}} \:,\:…\:\left({ii}\right) \\ $$$$\left({i}\right),\left({ii}\right)\Rightarrow{B}\mid{A} \\ $$$${if}\:\:“{n}''\:{is}\:\:{odd}\Rightarrow \\ $$$$\:{n}\mid\mathrm{1}^{{k}} +\left({n}−\mathrm{1}\right)^{{k}} \:,\:\mathrm{2}^{{k}} +\left({n}−\mathrm{2}\right)^{{k}} \:,\:…\:\left({iii}\right) \\ $$$$\:\frac{{n}+\mathrm{1}}{\mathrm{2}}\mid\mathrm{1}^{{k}} +{n}^{{k}} \:,\:\mathrm{2}^{{k}} +\left({n}−\mathrm{1}\right)^{{k}} \:,\:…\:\left({iv}\right) \\ $$$$\left({iii}\right),\left({iv}\right)\Rightarrow{B}\mid{A} \\ $$$$\Rightarrow\forall\:{n}\in\mathbb{N}\:\Rightarrow\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{n}\mid\:\mathrm{1}^{{k}} +\mathrm{2}^{{k}} +…+{n}^{{k}} \:\:\:;\:{if}\:\:“{k}''\:{is}\:\:“{odd}'' \\ $$$$ \\ $$