Solution

$($a$)$ Using the expression given for $K$ and $m,$ we get the following.

$K=m{c}^2-m_0{c}^2=\frac{m_0{c}^2}{\sqrt[\phantom{2}]{1-\frac{v^2}{c^2}}}-m_0{c}^2$

$=m_0{c}^2$

With $x=-\frac{v^2}{c^2},$ the Maclaurin series for $(1+x{)}^{-\frac{1}{2}}$ is most easily computed as a binomial series with $k=($Notice that $|x|<mn>1</mn>$ because $(1+x{)}^{-\frac{1}{2}}=1-$

$|+\frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}{x}^2-\frac{?}{3!}{x}^3+$cdots$||$