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Question Number 157061 by mathocean1 last updated on 19/Oct/21

Show that ∀ n ∈ N, ⌊((√n)+(√(n+1)))^2 ⌋=4n+1

$$ \\ $$$${Show}\:{that}\:\forall\:{n}\:\in\:\mathbb{N},\: \\ $$$$\lfloor\left(\sqrt{{n}}+\sqrt{{n}+\mathrm{1}}\right)^{\mathrm{2}} \rfloor=\mathrm{4}{n}+\mathrm{1} \\ $$

Answered by apriadodir last updated on 19/Oct/21

answer: ⌊((√n) + (√(n+1)) )^2 ⌋ = ⌊n+(n+1)+2((√(n(n+1))) )⌋ = 2n+1+2(√(n(n))) (because 2(√(n(n))) ≤ 2(√(n(n+1))) ) = 2n+1+2n = 4n +1

$$\mathrm{answer}: \\ $$$$\lfloor\left(\sqrt{\mathrm{n}}\:+\:\sqrt{\mathrm{n}+\mathrm{1}}\:\right)^{\mathrm{2}} \:\rfloor\:=\:\:\lfloor\mathrm{n}+\left(\mathrm{n}+\mathrm{1}\right)+\mathrm{2}\left(\sqrt{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}\:\right)\rfloor \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{2n}+\mathrm{1}+\mathrm{2}\sqrt{\mathrm{n}\left(\mathrm{n}\right)}\:\:\:\:\left(\mathrm{because}\:\mathrm{2}\sqrt{\mathrm{n}\left(\mathrm{n}\right)}\:\leqslant\:\mathrm{2}\sqrt{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{2n}+\mathrm{1}+\mathrm{2n} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{4n}\:+\mathrm{1} \\ $$

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

$$\left[{n}+{x}\right]={n}+\left[{x}\right] \\ $$$$\left[\left(\sqrt{{n}}+\sqrt{\mathrm{1}+{n}}\right)^{\mathrm{2}} \right]=\left[\mathrm{1}+\mathrm{2}{n}+\mathrm{2}\sqrt{{n}\left({n}+\mathrm{1}\right)}\right] \\ $$$$\left.=\mathrm{1}+\mathrm{2}{n}+\left[\mathrm{2}\sqrt{{n}\left(\mathrm{1}+{n}\right.}\right)\right]...\left({E}\right) \\ $$$${n}.{n}<{n}\left({n}+\mathrm{1}\right)<\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{2}\sqrt{{n}^{\mathrm{2}} }<\mathrm{2}\sqrt{{n}\left({n}+\mathrm{1}\right)}<\mathrm{2}\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}{n}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}{n}\leqslant\mathrm{2}\sqrt{{n}\left({n}+\mathrm{1}\right)}<\mathrm{2}{n}+\mathrm{1}\Rightarrow\left[\sqrt{{n}\left({n}+\mathrm{1}\right)}\right]=\mathrm{2}{n} \\ $$$$\Rightarrow\left[\sqrt{{n}}+\sqrt{\mathrm{1}+{n}}\right]^{\mathrm{2}} =\mathrm{4}{n}+\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

Commented by mathocean1 last updated on 19/Oct/21

Thanks guys.

$${Thanks}\:{guys}. \\ $$

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