Solution of the Recurrence is
$T(n) = \sqrt{n} + \left(\dfrac{n}{2} \right)$
$T(1) = 1$
$T(n) = T\left(\dfrac{n}{2}\right) + \left(n\right)^{\frac{1}{2}}$
$T(n) = \left[T\left(\dfrac{n}{2^{2}}\right) + \left(\dfrac{n}{2}\right)^{\frac{1}{2}}\right] + \left( n \right)^{\frac{1}{2}}$
$T(n) = T\left(\dfrac{n}{2^{3}}\right) + \left(\dfrac{n}{2^{2}}\right)^{\frac{1}{2}} + \left(\dfrac{n}{2}\right)^{\frac{1}{2}} + \left( n \right)^{\frac{1}{2}}$
$T(n) = T\left(\dfrac{n}{2^{k}}\right) + \left(\dfrac{n}{2^{k-1}}\right)^{\frac{1}{2}} + \left(\dfrac{n}{2^{k-2}}\right)^{\frac{1}{2}} + \left(\dfrac{n}{2^{k-3}}\right)^{\frac{1}{2}} + \cdot\cdot\cdot + \left( \dfrac{n}{2} \right)^{\frac{1}{2}} + \left(n \right)^{\frac{1}{2}} $
$\implies$ when $\dfrac{n}{2^{k}} = 1$
$\implies 2^{k} = n \implies k = \log_{2}n$
$T(n) = T(1) + \left(\dfrac{n}{2^{k}\cdot 2^{-1}}\right)^{\frac{1}{2}} + \left(\dfrac{n}{2^{k}\cdot2^{-2}}\right)^{\frac{1}{2}} + \left(\dfrac{n}{2^{k}\cdot2^{-3}}\right)^{\frac{1}{2}} + \cdot\cdot\cdot + \left( \dfrac{n}{2} \right)^{\frac{1}{2}} + \left(n \right)^{\frac{1}{2}} $
$T(n) = 1 + \left(2\right)^{\frac{1}{2}} + \left(4\right)^{\frac{1}{2}} + \left(8\right)^{\frac{1}{2}} + \cdot\cdot\cdot + \left(n\right)^{\frac{1}{2}}$
$T(n) = 1 + \left(2\right)^{\frac{1}{2}} + \left(2^{2}\right)^{\frac{1}{2}} + \left(2^{3}\right)^{\frac{1}{2}} + \cdot\cdot\cdot + \left(n\right)^{\frac{1}{2}}$
$T(n) = 2^{0} + \left(2\right)^{\frac{1}{2}} + 2 + \left(2 \right)^{\frac{3}{2}} + \cdot\cdot\cdot +\log_{2} n \ \ \text{times} + \sqrt{n}$
$T(n) = \dfrac{\left(\sqrt{2}\right)^{\log_{2}n} - 1}{\left(\sqrt{2} - 1\right)} + \sqrt{n}$
$T(n) = \dfrac{2 ^{\log_{2}n^{\frac{1}{2}}} - 1}{\sqrt{2} - 1} + \sqrt{n}$
$T(n) = \dfrac{\sqrt{n} - 1}{\sqrt{2} - 1} + \sqrt{n}$