Nested Square Root Sum (And Triangular Numbers)

$\begin{aligned} \sum_{k=1}^{2015} \sqrt{\sqrt{T_k+\dfrac{1}{8}}-\sqrt{T_k}} &= \sum_{k=1}^{2015} \sqrt{\sqrt{\dfrac{k(k+1)}{2}+\dfrac{1}{8}}-\sqrt{\dfrac{k(k+1)}{2}}} \\ &= \dfrac{1}{\sqrt[4]{2}}\sum_{k=1}^{2015} \sqrt{\sqrt{k(k+1)+\dfrac{1}{4}}-\sqrt{k(k+1)}} \\ &= \dfrac{1}{\sqrt[4]{2}}\sum_{k=1}^{2015} \sqrt{k+\dfrac{1}{2}-\sqrt{k(k+1)}} \\ &= \dfrac{1}{\sqrt[4]{2}}\sum_{k=1}^{2015} \sqrt{\left(\sqrt{\dfrac{k+1}{2}}-\sqrt{\dfrac{k}{2}}\right)^2} \\ &= \dfrac{1}{\sqrt[4]{2}}\sum_{k=1}^{2015} \sqrt{\dfrac{k+1}{2}}-\sqrt{\dfrac{k}{2}}\\ &= \dfrac{1}{\sqrt[4]{2}}\left(\sqrt{1008}-\sqrt{\dfrac{1}{2}}\right)\\ &= \dfrac{1}{\sqrt[4]{2}}\sqrt{\left(\sqrt{1008}-\sqrt{\dfrac{1}{2}}\right)^2}\\ &= \dfrac{1}{\sqrt[4]{2}}\sqrt{\dfrac{2017}{2}-\sqrt{2016}}\\ &= \sqrt{\dfrac{1}{\sqrt{2}}\left(\dfrac{2017}{2}-\sqrt{2016}\right)}\\ &= \sqrt{\dfrac{2017}{2\sqrt{2}}-\sqrt{1008}}\\ &= \sqrt{\sqrt{\dfrac{2017^2}{8}}-\sqrt{1008}} \end{aligned}$

Thus $(p,q,r)=\left(2017^2, 8, 1008\right)$ and $p+q+r\equiv 17^2+8+8\equiv \boxed{305}\pmod{1000}$

Calvin Lin these are the generalizations you asked Daniel for.

$(i) ~\displaystyle \sum_{k=1}^{a} \sqrt{\sqrt{T_k+\dfrac{1}{8}}-\sqrt{T_k}} \Longrightarrow \dfrac{\sqrt{a+1}-1}{\sqrt[4]{2^3}}$

$(ii) ~\displaystyle \sum_{k=1}^{a} \sqrt{\sqrt{T_k+\dfrac{1}{b}}-\sqrt{T_k}} \Longrightarrow \displaystyle \sum_{k=1}^{a} \sqrt{\sqrt{\dfrac{bn^2+b}{2b}+\dfrac{2}{2b}}-\sqrt{T_k}}$

$\displaystyle \sum_{k=1}^{a} \sqrt{\sqrt{\dfrac{bn^2+bn}{2b}+\dfrac{2}{2b}}-\sqrt{T_k}}$

This tells us that b must be even. Thus we can rewrite $b=2k$

$\displaystyle \sum_{k=1}^{a} \sqrt{\sqrt{\dfrac{kn^2+kn}{b}+\dfrac{1}{b}}-\sqrt{T_k}} \Longrightarrow kn^2+kn+1$ is a perfect square because if it's not.... It's just ugly.

We can rewrite this as a perfect square of the form $(tn+s)^2$ . Thus we have $s^2=1\rightarrow s=1$ . Logic tells us that s is greater than 0. Finally, we have

$2t=k$ and $t^2=k$

Clearly, the only solution here is $t=2,~k=4$ . Thus since $b=2k\Rightarrow b=8$