Floor function and hexagonal numbers

Question

While playing around with squares, I wondered about the sum of square roots of all natural numbers between two perfect squares(both inclusive). After taking the floor value of the expression for first few $n$ the numbers then obtained turned out to be $6,15,28,\ldots$. These are nothing but the hexagonal Numbers. That is, $$ \lfloor \sum \limits_{h=0}^{2n+1} \sqrt{n^2+h} \rfloor = H(n+1) $$ Proving it was not a bit of a task, I started with truncating the Taylor expansion of the square root function and bounding the summand as follows : $$n+\frac{h}{2n} -\frac{h^2}{8n^3} < \sqrt{n^2+h} < n + \frac{h}{2n}$$ Summing it all up and with little manipulation we get $$ \lfloor \sum \limits_{h=0}^{2n+1} \sqrt{n^2+h} \rfloor = 2n^2+3n +1 = (n+1)(2(n+1)-1) = H(n+1) $$ Although I'm still struggling to find a more intuitive way which directly relates the sum with the very numbers. Any ideas/suggestions are appreciated.