What is the $n$-th sequence element for the generating function $\frac{1}{(1-ax)^2}$?
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for e.g. for $\frac{1}{(1-ax)} = a^n$ or for $\frac{1}{(1-x)^2} = n+1$
generating function = $\frac{1}{(1-ax)^2}$
recurrence-relationsgenerating-functions
asked 2011-11-10
user id:18430
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You are looking at the derivative of $1/(1-ax)$ with respect to $x$, divided by $a$. Hence the $x^n$ term in this series is... – 2011-11-10
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(-2)/(1-ax)^2 ? – 2011-11-10
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Hmmm... The $x^n$ term in a series ought to be $a_nx^n$ for a given coefficient $a_n$, depending (possibly) on $n$ but **not on** $x$. Anyway, in the meantime you received full answers, hence, unless you wish to ask precise questions on specific steps of these, let us stop here. – 2011-11-10