Question Number 157061 by mathocean1 last updated on 19/Oct/21
$$ \\ $$$${Show}\:{that}\:\forall\:{n}\:\in\:\mathbb{N},\: \\ $$$$\lfloor\left(\sqrt{{n}}+\sqrt{{n}+\mathrm{1}}\right)^{\mathrm{2}} \rfloor=\mathrm{4}{n}+\mathrm{1} \\ $$
Answered by apriadodir last updated on 19/Oct/21
$$\mathrm{answer}: \\ $$$$\lfloor\left(\sqrt{\mathrm{n}}\:+\:\sqrt{\mathrm{n}+\mathrm{1}}\:\right)^{\mathrm{2}} \:\rfloor\:=\:\:\lfloor\mathrm{n}+\left(\mathrm{n}+\mathrm{1}\right)+\mathrm{2}\left(\sqrt{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}\:\right)\rfloor \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{2n}+\mathrm{1}+\mathrm{2}\sqrt{\mathrm{n}\left(\mathrm{n}\right)}\:\:\:\:\left(\mathrm{because}\:\mathrm{2}\sqrt{\mathrm{n}\left(\mathrm{n}\right)}\:\leqslant\:\mathrm{2}\sqrt{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{2n}+\mathrm{1}+\mathrm{2n} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{4n}\:+\mathrm{1} \\ $$
Answered by mindispower last updated on 19/Oct/21
![[n+x]=n+[x] [((√n)+(√(1+n)))^2 ]=[1+2n+2(√(n(n+1)))] =1+2n+[2(√(n(1+n)))]...(E) n.n<n(n+1)<(n+(1/2))(n+(1/2)) ⇒2(√n^2 )<2(√(n(n+1)))<2(n+(1/2))=2n+1 ⇒2n≤2(√(n(n+1)))<2n+1⇒[(√(n(n+1)))]=2n ⇒[(√n)+(√(1+n))]^2 =4n+1](https://www.tinkutara.com/question/Q157066.png)
$$\left[{n}+{x}\right]={n}+\left[{x}\right] \\ $$$$\left[\left(\sqrt{{n}}+\sqrt{\mathrm{1}+{n}}\right)^{\mathrm{2}} \right]=\left[\mathrm{1}+\mathrm{2}{n}+\mathrm{2}\sqrt{{n}\left({n}+\mathrm{1}\right)}\right] \\ $$$$\left.=\mathrm{1}+\mathrm{2}{n}+\left[\mathrm{2}\sqrt{{n}\left(\mathrm{1}+{n}\right.}\right)\right]…\left({E}\right) \\ $$$${n}.{n}<{n}\left({n}+\mathrm{1}\right)<\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{2}\sqrt{{n}^{\mathrm{2}} }<\mathrm{2}\sqrt{{n}\left({n}+\mathrm{1}\right)}<\mathrm{2}\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}{n}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}{n}\leqslant\mathrm{2}\sqrt{{n}\left({n}+\mathrm{1}\right)}<\mathrm{2}{n}+\mathrm{1}\Rightarrow\left[\sqrt{{n}\left({n}+\mathrm{1}\right)}\right]=\mathrm{2}{n} \\ $$$$\Rightarrow\left[\sqrt{{n}}+\sqrt{\mathrm{1}+{n}}\right]^{\mathrm{2}} =\mathrm{4}{n}+\mathrm{1} \\ $$$$ \\ $$$$ \\ $$
Commented by mathocean1 last updated on 19/Oct/21
$${Thanks}\:{guys}. \\ $$