May 28th, 2018, 09:54 PM  #1 
Newbie Joined: May 2018 From: Seoul Posts: 1 Thanks: 0  recurrence formula problem
how to solve it? a0=a1=1 thank you!! 
May 29th, 2018, 02:21 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,713 Thanks: 1806 
Why do you need to "solve" this? Have you realized that you can deal with the numerator and denominator separately? The denominator leads to $n$ factorial. 
May 29th, 2018, 12:23 PM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 474 Thanks: 262 Math Focus: Dynamical systems, analytic function theory, numerics 
What do you mean by "solve"? My guess from your screenshot is that you are solving a linear ODE by power series expansion. The resulting power series described by the recurrence IS the solution. What more could you ask for?

June 18th, 2018, 06:06 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
You have a formula that says that $\displaystyle a_{n+2}= \frac{3n+2}{(n+1)(n+2)}a_n$. You are also told that $\displaystyle a_0= a_1= 1$ so $\displaystyle a_2= a_{0+2}= \frac{3(0)+ 2}{(0+1)(0+2)}(1)= 1$, $\displaystyle a_3= a_{1+ 2}= \frac{3(1)+ 2}{(1+1)(1+ 2)}(1)= \frac{5}{6}$, $\displaystyle a_4= a_{2+ 2}= \frac{3(2)+ 1}{(2+ 1)(2+ 2)}(1)= \frac{7}{12}$, $\displaystyle a_5= a_{3+ 2}= \frac{3(3)+ 1}{(3+1)(3+ 2)}\left(\frac{5}{6}\right)= \frac{5}{12}$, etc. I don't see that giving any reasonable general formula. What is the exact statement of the problem? 

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