nice.. very nice please, give me a solution

Make the guess that $a_k-b_k=(a\sqrt k+b\sqrt{k+1}+c\sqrt{k-1})^2\\ =a^2k+b^2(k+1)+c^2(k-1)+2ab\sqrt{k^2+k}+2ac\sqrt{k^2-k}+2bc\sqrt{k^2-1}$. Comparing coefficients, we find that $ab=ac=-1$ and $bc=\tfrac12$, and $a=\sqrt2$, $b=c=\tfrac{1}{\sqrt2}$ works. So $\sqrt{a_k-b_k}=\sqrt{2k}-\sqrt{\tfrac{k+1}{2}}-\sqrt{\tfrac{k-1}{2}}$ (making sure to check that it actually works and is positive). Now, watch it telescope!

Zabelman, one question. You say "make the guess that"---but what would lead you to make such a guess in problems of this sort? Your solutions to problems in the forum often involve you saying something like "make the educated guess that" or "hypothesize so and so," but some people just aren't so incredibly smart that they can get flashes of insight like that (keep in mind you were the ONLY person in the U.S. to get 150 on the AMC 12A this year). What was your motivation in this problem? And can you give any tips, in general, on how to search for such seemingly random "guesses" for how to solve problems? Do you have some kind of system?

I do have a process: hope. The thing we're trying to evaluate involves $\sqrt{a_k-b_k}$, so I hope this comes out to something manageable. Next, the value of $a_k-b_k=3k+\sqrt{k+1}\sqrt{k-1}-2\sqrt{k}\sqrt{k+1}-2\sqrt{k}\sqrt{k-1}$ involved parwise products of $\sqrt{k}$, $\sqrt{k+1}$, and $\sqrt{k-1}$, hence the guess. Many problems are solved simply by hoping the right thing.

Here's what I did: First I factored, $b_n=2\sqrt{n}(\sqrt{n+1}+\sqrt{n-1})$. Then I noticed that $\sqrt{n+1}\sqrt{n-1}=\sqrt{n^{2}-1}$ which is part of $a_{n}$. So, I say $r=(\sqrt{n+1}+\sqrt{n-1})$ and square it to get $r^{2}=2n+2\sqrt{n^{2}-1}$. Then I said $\displaystyle a_n=\frac{4n+r^2}{2}$ and noticed $4n=(2\sqrt{n})^2=s^{2}$. From these it immediately follows that $\displaystyle a_n=\frac{s^2+r^2}{2}$, $b_n=rs$, and $\displaystyle a_n-b_n=\frac{s^2-2rs+r^2}{2}\Rightarrow \sqrt{a_n-b_n}=\frac{s-r}{\sqrt{2}}=\frac{2\sqrt{n}-(\sqrt{n+1}+\sqrt{n-1})}{\sqrt{2}}=\frac{(\sqrt{n}-\sqrt{n-1})-(\sqrt{n+1}-\sqrt{n})}{\sqrt{2}}$ So, $\displaystyle\sum^{49}_{n=1}\sqrt{a_n-b_n}=\displaystyle\sum^{49}_{n=1}\frac{(\sqrt{n}-\sqrt{n-1})-(\sqrt{n+1}-\sqrt{n})}{\sqrt{2}} \\ \displaystyle=\frac{1}{\sqrt{2}}\left(\sum^{49}_{n=1}(\sqrt{n}-\sqrt{n-1})-\sum^{49}_{n=1}(\sqrt{n+1}-\sqrt{n})\right) \\ \displaystyle=\frac{1}{\sqrt{2}}\left(\left(\sum^{49}_{n=1}\sqrt{n}-\sum^{49}_{n=1}\sqrt{n-1}\right)-\left(\sum^{49}_{n=1}\sqrt{n+1}-\sum^{49}_{n=1}\sqrt{n}\right)\right) \\ \displaystyle=\frac{1}{\sqrt{2}}\left(\left(\sum^{49}_{n=1}\sqrt{n}-\sum^{48}_{n=0}\sqrt{n}\right)-\left(\sum^{50}_{n=2}\sqrt{n}-\sum^{49}_{n=1}\sqrt{n}\right)\right) \\ \displaystyle=\frac{1}{\sqrt{2}}\left(\left(\sqrt{49}-\sqrt{0}\right)-\left(\sqrt{50}-\sqrt{1}\right)\right) \\ \displaystyle=-5+4\sqrt{2}$