Answer
$\left(\begin{array}{l} 30\\ 0 \end{array}\right)(x^{1/2})^{30}(1)^0=x^{15}$
$\left(\begin{array}{l}
30\\ 1 \end{array}\right)(x^{1/2})^{29}(1)^1=30x^{29/2}$
$\left(\begin{array}{l}30\\2\end{array}\right)(x^{1/2})^{28}(1)^{2}=435x^{14}$
$\left(\begin{array}{l}30\\3\end{array}\right)(x^{1/2})^{27}(1)^{3}=4060x^{27/2}$
Work Step by Step
We find the first 4 terms in the expansion of $(x^{1/2}+1)^{30}$ by using Binomial Theorem coefficients:
$\left(\begin{array}{l} 30\\ 0 \end{array}\right)(x^{1/2})^{30}(1)^0=x^{15}*1=x^{15}$
$\left(\begin{array}{l}
30\\ 1 \end{array}\right)(x^{1/2})^{29}(1)^1=30x^{29/2}$
$\left(\begin{array}{l}30\\2\end{array}\right)(x^{1/2})^{28}(1)^{2}=435x^{14}$
$\left(\begin{array}{l}30\\3\end{array}\right)(x^{1/2})^{27}(1)^{3}=4060x^{27/2}$
