Determine the sum of $12 + 12 + 12 + \cdots + 12$ (there are n+1 terms) by using OGF

Question

I bet all of this problem viewers know that the answer is $12 (n+1)$, but it wants us to use ordinary generating function to calculate the summation. I try to change the series into sequence $ (a_n) = (12, 24, 36, \cdots) $ and then substitute it to the P (x) as the OGF of the sequence, which $a_n = 12 (n+1)$ and that is it the coefficient of $x^n$ is the sum of the series.

But that seems doesn't right. So, do you have any idea?

Answer 1

A useful fact to know is that multiplication of a series $A(x)=\sum_{n=0}^\infty a_nx^n$ with the geometric series $\frac{1}{1-x}$ transforms the sequence $(a_n)_{n\geq 0}$ into a sequence of sums $\left(\sum_{k=0}^na_k\right)_{n\geq 0}$.

\begin{align*} \frac{1}{1-x}A(x)&=\frac{1}{1-x}\sum_{n=0}^\infty a_n x^n\\ &=\sum_{n=0}^\infty \left(\sum_{k=0}^n a_k\right) x^n\tag{1} \end{align*}

Here we start with \begin{align*} \frac{12}{1-x}&=12\sum_{n=0}^\infty x^n=\sum_{n=0}^\infty 12 x^n \end{align*}

We obtain thanks to (1) \begin{align*} \color{blue}{\frac{12}{(1-x)^2}} &=\sum_{n=0}^\infty\left(\sum_{k=0}^n 12\right) x^n\\ &\color{blue}{=\sum_{n=0}^\infty 12(n+1) x^n}\\ \end{align*}