If $\displaystyle A = \sum^{n^2-1}_{k=1}\sqrt{\sqrt{2n}+\sqrt{n+\sqrt{k}}}$ and $\displaystyle B = \sum^{n^2-1}_{k=1}\sqrt{\sqrt{2n}-\sqrt{n+\sqrt{k}}}$, Then $\displaystyle A\cdot B^{-1} = $
Try: Iam trying to convert $A+B$ into $A$ or in $B$ form like this way $$\bigg(\sqrt{\sqrt{2n}+\sqrt{n+\sqrt{k}}}+\sqrt{\sqrt{2n}-\sqrt{n+\sqrt{k}}}\bigg)^2=2\sqrt{2n}+2\sqrt{\sqrt{2n}-\sqrt{n+\sqrt{k}}}$$ $$\bigg(\sqrt{\sqrt{2n}+\sqrt{n+\sqrt{k}}}+\sqrt{\sqrt{2n}-\sqrt{n+\sqrt{k}}}\bigg)=\sqrt{2}\bigg(\sqrt{\sqrt{2n}+\sqrt{\sqrt{2n}-\sqrt{n+\sqrt{k}}}}\bigg)$$
could some help me to solve further, thanks
