$(\sqrt{n}+\sqrt{n+1})^2=2n+1+2\sqrt{n(n+1)}$ let's do some bounding, clearly $2n<2\sqrt{n(n+1)}<2n+1$ adding $2n+1$ we get, $4n+1<n+2\sqrt{n(n+1)}+n+1<4n+2\implies\sqrt{4n+1}<(n)+\sqrt{n}+\sqrt{n+1}<\sqrt{4n+2}$ since $[\sqrt{4n+1}]=\sqrt{4n+2}]\quad\forall n\in\mathbb{N}$ we obtain the desired equality.$\blacksquare$