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Question Number 192425 by mehdee42 last updated on 17/May/23

$$Question$$ $$if\mathrm{♮}k\epsilon isodd\mathrm{\&}A=1^k+2^k+...+n^k\mathrm{\&}B=1+2+...+n$$ $$provethat:B\mid A$$

Answered by MM42 last updated on 24/May/23

$$Wekhnow;1+2+...+n=\frac{n(n+1)}{2}\mathrm{\&}(n,\frac{n+1}{2})=1or(\frac{n}{2},n+1)=1$$ $$Thus,itsofficestoshowthat$$ $$n\mid A\mathrm{\&}\frac{n+1}{2}\mid Aor\frac{n}{2}\mid A\mathrm{\&}n+1\mid A$$ $$proof\therefore$$ $$if\mathrm{♮}n\epsilon iseven\Rightarrow$$ $$\frac{n}{2}\mid 1^k+{(n-1)}^k,2^k+{(n-2)}^k,...\left(i\right)$$ $$n+1\mid 1^k+n^k,2^k+{(n-1)}^k,...\left(ii\right)$$ $$\left(i\right),\left(ii\right)\Rightarrow B\mid A$$ $$if\mathrm{♮}n\epsilon isodd\Rightarrow$$ $$n\mid 1^k+{(n-1)}^k,2^k+{(n-2)}^k,...\left(iii\right)$$ $$\frac{n+1}{2}\mid 1^k+n^k,2^k+{(n-1)}^k,...\left(iv\right)$$ $$\left(iii\right),\left(iv\right)\Rightarrow B\mid A$$ $$\Rightarrow \mathrm{\forall }n\in \mathbb{N}\Rightarrow 1+2+3+...+n\mid 1^k+2^k+...+n^k;if\mathrm{♮}k\epsilon is\mathrm{♮}odd\epsilon$$ $$$$

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