Given a function f(x)=x², define g₁(x)=f(x) and g_((n+1)) (x)=min_(0≤t≤x) (g_n (t)+f(x-t)), where n∈N, then

 $g_2\left(x\right)=\underset{0\le t\le x}{\text{min}}\left(g_1\left(t\right)+f(x-t)\right)=\underset{0\le t\le x}{\text{min}}(t^2+{(x-t)}^2)=\underset{0\le t\le x}{\text{min}}2({(t-\frac{x}{2})}^2+\frac{x^2}{4})=\frac{x^2}{2}$
Similarly,  $g_3\left(x\right)=\frac{x^2}{3}\dots \mathrm{..};g_n\left(x\right)=\frac{x^2}{n}$
$\Rightarrow \underset{n=1}{\overset{2024}{ }}{g}_n\left(\frac{1}{\sqrt{n+1}}\right)=\underset{n=1}{\overset{2024}{ }}\frac{1}{n}-\frac{1}{n+1}=\frac{2024}{2025} and \underset{n=2}{\overset{2024}{ }}{g}_n\left(\frac{1}{\sqrt{n-1}}\right)=\underset{n=2}{\overset{2024}{ }}\frac{1}{n-1}-\frac{1}{n}$$=\frac{2023}{2024}$
                                             

Similarly,  $g_3\left(x\right)=\frac{x^2}{3}\dots \mathrm{..};g_n\left(x\right)=\frac{x^2}{n}$
 
     $\Rightarrow \underset{n=1}{\overset{2024}{ }}{g}_n\left(\frac{1}{\sqrt{n+1}}\right)=\underset{n=1}{\overset{2024}{ }}\frac{1}{n}-\frac{1}{n+1}=\frac{2024}{2025} and \underset{n=2}{\overset{2024}{ }}{g}_n\left(\frac{1}{\sqrt{n-1}}\right)=\underset{n=2}{\overset{2024}{ }}\frac{1}{n-1}-\frac{1}{n}$$=\frac{2023}{2024}$