+289 Simplify \[\frac{\sqrt{45 + \sqrt{1}} + \sqrt{45 + \sqrt{2}} + \sqrt{45 + \sqrt{3}} + \dots + \sqrt{45 + \sqrt{2024}}}{\sqrt{45 - \sqrt{1}} + \sqrt{45 - \sqrt{2}} + \sqrt{45 - \sqrt{3}} + \dots + \sqrt{45 - \sqrt{2024}}}.\]
+9673 We let \(A = \sqrt{45 + \sqrt1} + \sqrt{45 + \sqrt2} + \cdots + \sqrt{45 + \sqrt{2024}} = \displaystyle\sum_{k = 1}^{2024} \sqrt{45 + \sqrt k}\) and \(B = \sqrt{45 - \sqrt 1} + \sqrt{45 - \sqrt2} + \cdots + \sqrt{45 - \sqrt{2024}} = \displaystyle\sum_{k = 1}^{2024} \sqrt{45 - \sqrt{k}}\), and then consider A - B.
Now, \(A -B = \displaystyle\sum_{k = 1}^{2024} \left(\sqrt{45 + \sqrt k} - \sqrt{45 - \sqrt k}\right)\). But what is the summand?
If we square it, that is \(\left(\sqrt{45 + \sqrt k} - \sqrt{45 - \sqrt k}\right)^2 = (45 + \sqrt k) - 2\sqrt{(45 - \sqrt k)(45 + \sqrt k)} + (45 - \sqrt k) = 90 - 2 \sqrt{2025 - k}\).
Taking square root gives \(\sqrt{45 + \sqrt k} - \sqrt{45 - \sqrt k} = \sqrt 2 \cdot \sqrt{45 - \sqrt{2025 - k}}\).
Therefore, \(A - B = \sqrt 2\displaystyle\sum_{k = 1}^{2024}\sqrt{45 - \sqrt{2025 - k}}\). Making the substitution n = 2025 - k and replacing the dummy variable n by k gives \(A - B= \sqrt 2 \displaystyle\sum_{k = 1}^{2024} \sqrt{45 - \sqrt k} = B \sqrt 2\).
That means \(B (1 + \sqrt 2) = A\), and therefore \(\begin{array}{rcl} \dfrac{\sqrt{45 + \sqrt{1}} + \sqrt{45 + \sqrt{2}} + \sqrt{45 + \sqrt{3}} + \dots + \sqrt{45 + \sqrt{2024}}}{\sqrt{45 - \sqrt{1}} + \sqrt{45 - \sqrt{2}} + \sqrt{45 - \sqrt{3}} + \dots + \sqrt{45 - \sqrt{2024}}} &=& \dfrac AB\\ &=& \dfrac{(1 + \sqrt 2)B}B\\ &=& \boxed{1 + \sqrt 2} \end{array}\)
In fact, you can prove that:
\(\displaystyle \dfrac{\displaystyle \sum_{k = 1}^{n^2-1}\sqrt{n^2 - \sqrt k}}{\displaystyle \sum_{k = 1}^{n^2-1}\sqrt{n^2 + \sqrt k}} = 1 + \sqrt 2\) for any positive integer n > 1.
