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

$$$$$$Showthat\mathrm{\forall }n\in \mathbb{N},$$$$\lfloor {(\sqrt{n}+\sqrt{n+1})}^2\rfloor =4n+1$$

Answered by apriadodir last updated on 19/Oct/21

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

Answered by mindispower last updated on 19/Oct/21

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

Commented by mathocean1 last updated on 19/Oct/21

$$Thanksguys.$$

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