A while back, a problem asked me to simplify the nested radical$$\dfrac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\dots+\sqrt{10+\sqrt{98}}+\sqrt{10+\sqrt{99}}}{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\dots+\sqrt{10-\sqrt{98}}+\sqrt{10-\sqrt{99}}}=\frac{\sum_{k=1}^{99}\sqrt{10-\sqrt{k}}}{\sum_{k=1}^{99}\sqrt{10+\sqrt{k}}}$$
This problem can be simply solved using the idea that $$\sqrt{m-\sqrt{n}}+\sqrt{m+\sqrt{n}}=\sqrt{2}\times \sqrt{m+\sqrt{m^2-n}}$$
Which implies that $$\sum_{k=1}^{99}\sqrt{10-\sqrt{k}}+\sum_{k=1}^{99}\sqrt{10+\sqrt{k}}=\sqrt{2}\sum_{k=1}^{99}\sqrt{10+\sqrt{100-k}}$$ Thus we get that $$\sum_{k=1}^{99}\sqrt{10-\sqrt{k}}=(\sqrt{2}-1)\times \sum_{k=1}^{99}\sqrt{10+\sqrt{k}}$$
Thus the answer is $\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$.
However, it took me a long time to find this solution. In truth, the solution above seems rather lucky.
Also, it was actually hard to believe the value was $\sqrt{2}+1$, which seemed to have nothing to do with the number $10$ itself.
Is there a more intuitive/alternate approach to this problem?
