# Chapter 8, Sequences and Series - Section 8.6 - The Binomial Theorem - 8.6 Exercises - Page 636: 30

$\left(\begin{array}{l} 30\\ 0 \end{array}\right)(x^{1/2})^{30}(1)^0=x^{15}$ $\left(\begin{array}{l} 30\\ 1 \end{array}\right)(x^{1/2})^{29}(1)^1=30x^{29/2}$ $\left(\begin{array}{l}30\\2\end{array}\right)(x^{1/2})^{28}(1)^{2}=435x^{14}$ $\left(\begin{array}{l}30\\3\end{array}\right)(x^{1/2})^{27}(1)^{3}=4060x^{27/2}$

#### Work Step by Step

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

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.