Find the closed form of the generating function

Question

Write the generating function in closed form (finite fraction): $$a(z) = 2 + 4z + 6z^2 + 8z^3 + 10z^4 + \dots $$

I know that $\sum_{i = 0}^{\infty}{2(i+1)z^i}$ is the generating function in summation form but I'm not sure how to represent that as a finite fraction. In my course, we are using the proposition:

$\frac{1}{1-(\alpha)z} = \alpha^0z^0 + \alpha^1z^1+ \alpha^2z^2 + \dots + \alpha^iz^i\dots $

Answer 1

Notice that your sum is equivalent to $2 \sum_{n=0}^\infty nz^n$. Then consider the geometric series:

$$\sum_{n=0}^\infty z^n = \frac{1}{1-z}$$

(whenever $|z|<1$). Take the derivative of both sides with respect to $z$, then multiply both sides by $z$. You'll find an expression for the summation you have. I'll leave the calculations/justification up to you.

Answer 2

By the binomial theorem,$$\begin{align}(1-z)^{-2}&=\sum_{i\ge0}\binom{-2}{i}(-z)^i\\&=\sum_{i\ge0}\frac{(-2)\cdots(-(i+1))}{i!}(-z)^i\\&=\sum_i(i+1)z^i,\end{align}$$so $a(z)=\frac{2}{(1-z)^2}$.