Solving Recurrence Relation: a_n=\frac{a_{n-5}}{n(n-1)}

[naes213]

Homework Statement

Got this recurrence relation when trying to solve for a series solution to a differential equation: $$a_n=\frac{a_{n-5}}{n(n-1)} , a_0,a_1=constant , a_2,a_3,a_4=0$$

Homework Equations

The Attempt at a Solution

My attempt at a solution involved first writing out all the terms which led to the pattern $$a_5=\frac{a_0}{5\cdot4}$$ and $$a_6=\frac{a_1}{6\cdot5}$$ and $$a_{10}=\frac{a_0}{10\cdot9\cdot5\cdot4}...a_{15}=\frac{a_0}{15\cdot14\cdot10\cdot9\cdot5\cdot4}$$

I ended up with $$a_n=\frac{a_0}{5^n \cdot n!}$$ missing something on the bottom which as far as I can tell is $$(5(n)-1)(5(n-1)-1)(5(n-2)-1)...$$
The problem is that i can't figure out a more concise simple expression for this last part. Any help would be greatly appreciated!

 

[mighty2000]

Thank you for sharing your attempt at solving this recurrence relation. It looks like you have made good progress in finding a pattern and a general formula for the terms. However, as you have mentioned, there seems to be a factor missing in the denominator.

In order to find a more concise expression for this last factor, I would suggest looking at the pattern of the terms in the numerator. It seems that each term in the numerator is decreasing by 5, starting with a_0 and then a_1, a_2, a_3, etc. Can you think of a way to incorporate this pattern into the denominator? Maybe by using a factorial or a similar operation?

Another approach could be to look at the original differential equation and see if there are any properties or relationships that could help in finding a more concise expression for the denominator. It might also be helpful to consult with your classmates or professor for any tips or insights.

I hope this helps and good luck with your solution! Keep up the good work in tackling this challenging problem.