I toyed with common series a bit and came up with
$\sum \limits_{k=0}^\infty ~(-1)^k x^{3k-1} = \dfrac{x^2}{1+x^3}$
Look at the common Maclaurin series templates.
You may have to use a power of $x$ rather than just $x$
You may have to adjust things by dividing or multiplying by $x$
Use Wolfram Alpha Series function to test out your hypotheses.
Series[f(x), {x, 0, n}]
will give you $n$ terms of the series of $f(x)$ about $x=0$
$\displaystyle \sum_{n=0}^\infty x^{3n+2}= \sum_{n=0}^\infty (x^3)^nx^2= x^2\sum_{n=0}^\infty (x^3)^n$ which is now a geometric series in $\displaystyle x^3$. Since $\displaystyle \sum_{n= 0}^\infty r^n= \frac{1}{1- r}$ the sum of this series is $\displaystyle x^2\frac{1}{1- x^3}= \frac{x^2}{1- x^3}$.