Expressing the generating function defined by $b_n = \sum_{k=0}^{n} 3^k\cdot a_k$

Question

The title is probably somewhat unclear, sorry if it is..

Let $F$ be the generating function of the sequence $(a_n)_{n=0}^{\infty}$

Use $F$ to express the generating function for $(b_n)_{n=0}^{\infty}$ that is defined by $b_n = \sum_{k=0}^{n} 3^k\cdot a_k$ .

I have went through my notes on generating functions, and most of it was about using this technique for solving some combinatorical problems. I would be happy to have some explanation on how to approach the above problem and other problems of this sort.

Thanks!

Answer 1

Recall that $$ \frac{3}{3-X} = \sum_{n=0}^\infty 3^{-n} X^n. $$ The product $\frac{3F(X)}{3-X}$ can be written, using the properties of Cauchy product, as $$ \frac{3F(X)}{3-X} = \left(\sum_{n=0}^\infty a_n X^n\right)\left(\sum_{n=0}^\infty 3^{-n} X^n\right) = \sum_{n=0}^\infty c_n X^n $$ with $c_n=\sum_{k=0}^n a_k 3^{-(n-k)} = 3^{-n} b_n$. Let us write $G$ for the generating function defined by $$ G(X) = \sum_{n=0}^\infty b_n X^n $$ Then $$G\!\left(\frac{X}{3}\right) = \sum_{n=0}^\infty 3^{-n}b_n X^n = \sum_{n=0}^\infty c_n X^n = \frac{3F(X)}{3-X}.$$ (or, equivalently, $G(X) = \frac{F(3X)}{1-X}$)