Find closed form of the ordinary generating function for sequences
Solution 1
Your second sequence as a generating function is $$ z^3+3z^4+9z^5+\dotsb=z^3(1+3z+9z^2+\dotsb)=\frac{z^3}{13z}. $$ while your first is $$ 2z+2z^5+2z^9+\dotsb=2z(1+z^4+z^8+\dotsb)=\frac{2z}{1z^4}. $$ These answers are based on the fact that your indexing starts at $1$.
Solution 2
$0,\; 0,\; 1,\; 3,\; 9,\; 27,\; 81,\; 243,\;\ldots$
its generating function is
$0+0x+1x^2+3x^3+9x^4+27x^5+\ldots$
$x^2(1+3x+9x^2+27x^3+\ldots)$
$$x^2 \sum_{n=0}^\infty (3x)^n=x^2\frac{1}{13x}=\color{red}{\frac{x^2}{13x}}$$
The first sequence has generating function
$$2+2x^4+2x^8+2x^{12}+\ldots=2\sum_{n=0}^\infty (x^4)^n=\color{red}{\frac{2}{1x^4}}$$
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gbm0102
Updated on October 30, 2020Comments

gbm0102 about 3 years
I'm trying to find the closed form of the ordinary generating function for the following sequences:
(1) $2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0,\dotsc.$
(2) $0, 0, 1, 3, 9, 27, 81, 243,\dotsc$
Here is my work. I think I've figured out the sequences. My problem is, how do I get these into closed form? From class, I'm used to seeing closed form in other ways; for example, the geometric series would be $1/(1x)$; binomial theorem would be $(1+z)^n$. But how would I write these in closed form? Also, am I correct in thinking that these numbers are meant to represent coefficients?

gbm0102 about 6 yearsQuick question: Could you explain where (x4)^2 came from in the summation? Would it not just be x^4?

Raffaele about 6 years@GarrettMcClure It was a typo. It was $(x^4)^n$ I have fixed it. Thank you...