0
$\begingroup$

How to find the sequence that is generated by this GF?

$B(x)=(x+3)^2 + \frac{x}{(1-3x)^6}$

We know that $\frac{1}{(1-ax)}$ is generated by $\sum_{i=0}^n a^n x^n$

$\endgroup$
1
$\begingroup$

You should remember that $$ \frac{1}{(1-z)^m} = \sum_{n=0}^\infty \left( m+n-1\atop n\right) z^n $$ This can be proven by using induction and the fact that $$ \sum_{k=0}^n \left( m+k\atop k \right) = \left( m+n+1 \atop n\right) $$

$\endgroup$
  • $\begingroup$ that's $1/(1+z)^m$ ! $\endgroup$ – G Cab Feb 11 at 9:47
  • $\begingroup$ This is what I was missing. Thank you so much! $\endgroup$ – Dimitris Prasakis Feb 11 at 10:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.