# Recurrence relation in terms of another sequence

How do I solve a recurrence of the form $$nd^{n-1}a_n+a_{n+1}d^{n+1}=b_n$$ for $a_n$, where $b_n$ is another (known) sequence and $d$ is a constant? My only idea was to use a generating function and try to separate the added terms on the left and hammer them into a form where they can go together, but I can't make it work.

• what is the objective ? a definition for $a_n$ ? – Cardinal Jul 16 '15 at 19:27
• Rewriting this we get $a_{n+1}= a_n \frac{-n}{d^2} + \frac{b_n}{d^{n+1}}$ – Jan Jul 16 '15 at 19:32
• @Cardinal Yes, in terms of $d$ and $b_n$. – Laertes Jul 16 '15 at 19:32

## 1 Answer

I guess that you would like to express $a_n$ in terms of $b_n$.

From $$a_{n+1}d^{n+1}+nd^{n-1}a_n=b_n \tag1$$you have, assuming $d \neq0$, $$a_{n+1}+nd^{-2}a_n=d^{-n-1}b_n \tag2$$ and you obtain $$\frac{(-1)^n}{\prod_{k=1}^n(kd^{-2})}a_{n+1}-\frac{(-1)^{n-1}nd^{-2}}{\prod_{k=1}^n(kd^{-2})}a_n=\frac{(-1)^n}{\prod_{k=1}^n(kd^{-2})}d^{-n-1}b_n \tag3$$ giving $$\frac{(-1)^n}{n!\:d^{-2n}}a_{n+1}-\frac{(-1)^{n-1}}{(n-1)!\:d^{-2(n-1)}}a_n=\frac{(-1)^n}{n!}d^{n-1}b_n. \tag4$$ Then, summing from $n=1$ to $n=N$, you get by telescoping, $$\frac{(-1)^N}{N!\:d^{-2N}}a_{N+1}-a_1=\sum_{n=1}^N\frac{(-1)^n}{n!}d^{n-1}b_n \tag5$$ and

$$a_{N+1}=(-1)^N \frac{N!}{d^{2N}}\sum_{n=1}^N\frac{(-1)^n}{n!}d^{n-1}b_n +(-1)^N \frac{N!}{d^{2N}}a_1. \tag6$$