Oh! Nice identity

Let $S=\sqrt{45+\sqrt{1}}+\sqrt{45+\sqrt{2}}+\sqrt{45+\sqrt{3}}+\ldots+\sqrt{45+\sqrt{2024}}$ and $T=\sqrt{45-\sqrt{1}}+\sqrt{45-\sqrt{2}}+\sqrt{45-\sqrt{3}}+\ldots+\sqrt{45-\sqrt{2024}}$

We have:

$\displaystyle S+T=\sum_{i=1}^{2024}\left(\sqrt{45+\sqrt{2025-i}}+\sqrt{45-\sqrt{2025-i}}\right)\\ \qquad\;\;\displaystyle=\sum_{i=1}^{2024}\sqrt{\left(\sqrt{45+\sqrt{2025-i}}+\sqrt{45-\sqrt{2025-i}}\right)^2}\\ \qquad\;\;\displaystyle =\sum_{i=1}^{2024}\sqrt{90+2\sqrt{i}}=\sqrt{2}S$

Thus, $T=(\sqrt{2}-1)S$ or $P=\dfrac{S}{T}=\dfrac{1}{\sqrt{2}-1}=\sqrt{2}+1\approx\boxed{2.414}$

I got my answer from wolframapha with:

(summation sqrt(45+sqrt(i)),i=1 to 2024)/(summation sqrt(45-sqrt(i)),i=1 to 2024)

We can start by thinking: what would happen if we subtracted $\sqrt{45-\sqrt{x}}$ from $\sqrt{45+\sqrt{x}}$ ? Knowing what happens might be useful in some way. Let: $y=\sqrt{45+\sqrt{x}}-\sqrt{45-\sqrt{x}}$ , so $\sqrt{45-\sqrt{x}}+y=\sqrt{45+\sqrt{x}}$ .

Then: $y^2=(45+\sqrt{x})+(45 -\sqrt{x})+2\sqrt{(45+\sqrt{x})(45-\sqrt{x})}$ .

Simplifying: $y^2=90-2\sqrt{2025-x}$ , so $y=\sqrt{90-2\sqrt{2025-x}}$ .

That means $\sqrt{45-\sqrt{x}}+\sqrt{90-2\sqrt{2025-x}}=\sqrt{45+\sqrt{x}}$ .

We can already see that it is getting useful. Now, back to the problem. We can use our identity here:

$\frac{\sqrt{45+\sqrt{1}}+\sqrt{45+\sqrt{2}}+\sqrt{45+\sqrt{3}}+\ldots+\sqrt{45+\sqrt{2024}}}{\sqrt{45-\sqrt{1}}+\sqrt{45-\sqrt{2}}+\sqrt{45-\sqrt{3}}+\ldots+\sqrt{45-\sqrt{2024}}}$ $= \frac{\displaystyle \sum_{x=1}^{2024} \sqrt{45+\sqrt{x}}}{\displaystyle \sum_{x=1}^{2024} \sqrt{45-\sqrt{x}}}$

$= \frac{\displaystyle \sum_{x=1}^{2024} \sqrt{45-\sqrt{x}}+\displaystyle \sum_{x=1}^{2024} \sqrt{90-2\sqrt{2025-x}}}{\displaystyle \sum_{x=1}^{2024} \sqrt{45-\sqrt{x}}}$ $= 1 + \frac{\displaystyle \sum_{x=1}^{2024} \sqrt{90-2\sqrt{2025-x}}}{\displaystyle \sum_{x=1}^{2024} \sqrt{45-\sqrt{x}}}$ .

The numerator can be reordered and the sum can be 'reversed':

$= 1 + \frac{\displaystyle \sum_{x=1}^{2024} \sqrt{90-2\sqrt{x}}}{\displaystyle \sum_{x=1}^{2024} \sqrt{45-\sqrt{x}}}$ . Now, it's getting interesting:

$= 1 + \frac{\displaystyle \sum_{x=1}^{2024} \sqrt{2}\sqrt{45-\sqrt{x}}}{\displaystyle \sum_{x=1}^{2024} \sqrt{45-\sqrt{x}}}$ . Cancelling:

$= 1 + \sqrt{2} \approx\ \boxed{2.414}$

Woow too good

write this code in java you will get the answer : double d=0.0,e=0.0; for(int i=1;i<=2024;i++) { d = d + (Math.sqrt(45 + Math.sqrt(i))); e = e + (Math.sqrt(45 - Math.sqrt(i))); } System.out.println("Final ans is :"+d/e);