Maths Question:
$\begin{align} & \text{Expand }{{(1+3x)}^{-\tfrac{1}{2}}} \\ & \text{Giving the first four terms and stating the series is valid} \\\end{align}$
Maths Solution:
$\begin{align} & {{(1+3x)}^{-\tfrac{1}{2}}}=1+\left( -\frac{1}{2} \right)(3x)+\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)}{2!}{{(3x)}^{2}}+\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)}{3!}{{(3x)}^{3}}+---- \\ & {{(1+3x)}^{-\tfrac{1}{2}}}=1-\frac{3x}{2}+\frac{3}{8}{{(9x)}^{2}}-\frac{15}{48}(27{{x}^{3}})+----- \\ & {{(1+3x)}^{-\tfrac{1}{2}}}=1-\frac{3x}{2}+\frac{27{{x}^{2}}}{8}-\frac{135{{x}^{3}}}{16}+--- \\ & \text{The range is valid within }-\tfrac{1}{3}<x<\tfrac{1}{3} \\\end{align}$
University mathstopic: