Given ,
$T(n)=T(n-1)+\sqrt{n} ;n\geq 1$
$T(n)=T(n-2)+\sqrt{n-1}+\sqrt{n}$
$T(n)=T(n-3)+\sqrt{n-2}+\sqrt{n-1}+\sqrt{n}$
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$T(n)=T(1)+\sqrt{2}+............+\sqrt{n-2}+\sqrt{n-1}+\sqrt{n}$
$T(n)=T(0)+\sqrt{1}+\sqrt{2}+............+\sqrt{n-2}+\sqrt{n-1}+\sqrt{n}$
$T(n)=\sqrt{1}+\sqrt{2}+............+\sqrt{n-2}+\sqrt{n-1}+\sqrt{n}$ as $T(0)=0$
Now this summation we can write like ,
$T(n)=\sum_{i=1}^{n}\sqrt{n}$ which is approximately can be written
$\sum_{i=1}^{n}\sqrt{n}$ $\approx$ $\int_{1}^{n}\sqrt{x} dx$
$\large \int_{1}^{n}\sqrt{x} dx$
=$\large \frac{2}{3}\left [ n^{\frac{3}{2}}-1 \right ]$
So,$T(n)$ can be closely approximate to $\large n^{\frac{3}{2}}$.
So, $\large T(n)=\Theta (n^{\frac{3}{2}})$ .