jmc

Notice that $\sqrt{n+\sqrt{n^2-1}}=\frac{1}{\sqrt{2}}\sqrt{2n+2\sqrt{(n+1)(n-1)}}=\frac{1}{\sqrt{2}}\left(\sqrt{n+1}+\sqrt{n-1}\right)$. Thus, we have $$\sum _{n=1}^{9800}\frac{1}{\sqrt{n+\sqrt{n^2-1}}}$$$$=\sqrt{2}\sum _{n=1}^{9800}\frac{1}{\sqrt{n+1}+\sqrt{n-1}}$$$$=\frac{1}{\sqrt{2}}\sum _{n=1}^{9800}\left(\sqrt{n+1}-\sqrt{n-1}\right)$$ This is a telescoping series; note that when we expand the summation, all of the intermediary terms cancel, leaving us with $\frac{1}{\sqrt{2}}\left(\sqrt{9801}+\sqrt{9800}-\sqrt{1}-\sqrt{0}\right)=70+49\sqrt{2}$, and $p+q+r=\boxed{121}$.