# Thread: Need help with Maclaurin Series problem

1. ## Need help with Maclaurin Series problem

How can i recognize the function?

If the first problem (a) is

f(x) = summation of x^(3n+2) startinng from n=0, what is the function then? I know that x^n is geometric series function, but im unable to conclude the function

2. ## Re: Need help with Maclaurin Series problem

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 $x=0$

3. ## Re: Need help with Maclaurin Series problem

Hello, $\displaystyle\sum_{k=0}^{\infty}(-1)^k x^{3k+2}=\frac{x^2}{1+x^3}$. Error in your post was rectified

4. ## Re: Need help with Maclaurin Series problem

$\displaystyle \sum_{n=0}^\infty x^{3n+2}= \sum_{n=0}^\infty (x^3)^nx^2= x^2\sum_{n=0}^\infty (x^3)^n$ 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 $\displaystyle x^2\frac{1}{1- x^3}= \frac{x^2}{1- x^3}$.

5. ## Re: Need help with Maclaurin Series problem

Hello, in the summation,you forgot to enter $(-1)^n$.Then you will get $\frac{x^2}{1+x^3}$ posted in #2 of this thread.