Summation of square roots?

Let c = $1/\sqrt{2}.$ Then $\begin{aligned} (2c\sqrt{n} - c\sqrt{n+1} - c\sqrt{n-1})^2 &= 4c^2n + c^2(n+1) + c^2(n-1) +2c^2\sqrt{n^2-1} - 4c^2\sqrt{n^2+n} - 4c^2\sqrt{n^2-n} \\ &= 3n + \sqrt{n^2-1} - 2(\sqrt{n^2+n}-\sqrt{n^2-n}) \\ &= a_n - b_n. \end{aligned}$ Note that the term in parentheses above is in fact positive (e.g. because the graph of $y=\sqrt{x}$ is concave down), so it's the positive square root of $a_n - b_n.$

Now, let $S = \sum_{k=1}^{49} \sqrt{k}.$ Then our sum is $\begin{aligned} \sum_{k=1}^{49} (2c\sqrt{k} -c\sqrt{k+1} -c\sqrt{k-1}) &= 2cS - c(S+\sqrt{50} - 1) - c(S - 7) \\ &= 8c - c\sqrt{50} \\ &= 4\sqrt{2} - 5, \end{aligned}$ so the answer is $\fbox{40}.$

Since all i had did so far with $LaTeX$ were $\Delta, \angle \cap \dots$ cope up with my latex abilities.

$a_n-b_n = 3n+ \sqrt{n^2-1} -2(\sqrt{n^{2}+n} - \sqrt{n^{2}-n}) = 2n+ \large(\frac{(n+1)}{\sqrt{2}}+ \frac{(n-1)}{\sqrt{2}})$ + $2 \frac{\sqrt{n^2-1}}{2} -2\sqrt{2n} \dfrac{\sqrt{n+1}+\sqrt{n-1}}{\sqrt{2}} = \large[\sqrt{2k}- \dfrac{\sqrt{k+1}+\sqrt{k-1}}{2}]^2$ Now you k all that cancellation and drama.