Rahul's confusing square root

$\color{#D61F06}{1+3+5+7+\dotsm+(2n-1)=n^2=\left(\frac{1+(2n-1)}{2}\right)^2}$ So $\color{#20A900}{\sqrt{1+3+5+\dotsm+99}=\sqrt{\left(\frac{1+99}{2}\right)^2}=\sqrt{50^2}=\boxed{50}}$

We know that sum of $N$ odd numbers is $N^2$
Here $N = \frac{(99 + 1)} {2} = 50\\ 1 + 3 + ..... + 99 = 50^2\\ \sqrt{1 + 3 + .... + 99} = 50$