# Solving a recurrence relation, $a_n = \sqrt{n(n+1)}a_{n-1} + n!(n+1)^{3/2}$

I'm trying to solve the following recurrence relation, but I have a problem with the factorial part. I would like to evaluate its particular solution. I would like also to suggest a textbook for solving these kinds of questions. We have:

$$a_n = \sqrt{n(n+1)}a_{n-1} + n!(n+1)^{\frac{3}{2}}$$

• Hi @AhmedSaleh, I've formated your equation with latex so that possible future edits become easier. I encourage you to learn the basics of typing math with latex so that you can get the best out of Math.SE! Good luck! – user1620696 May 16 '13 at 23:13

OK, here's my solution. Note you didn't give any boundary values. Your equation can be rewritten as $$\frac{a_n}{n!}=\frac{a_{n-1}\sqrt{n(n+1)}}{n!}+(n+1)^{\frac{3}{2}}$$ Denote $b_n=\frac{a_n}{n!}$: $$b_n=b_{n-1}\sqrt{\frac{n+1}{n}}+(n+1)^{\frac{3}{2}}\\ \frac{b_n}{\sqrt{n+1}}=\frac{b_{n-1}}{\sqrt{n}}+(n+1)^3$$ Denote $s_n=\frac{b_{n}}{\sqrt{n+1}}$: $$s_n=s_{n-1}+(n+1)^3=\ldots=s_0+S_n$$ here $$S_n=\sum_{k=1}^{n+1}k^3=O(n^4)$$ I don't give the calculation of this expression, as you can easily find it by methods such as perturbation, induction, etc. Hence your expression is $$a_n=n!\sqrt{n+1}(a_0+S_n)$$