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May 28th, 2018, 09:54 PM   #1
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recurrence formula problem

how to solve it?

a0=a1=1

thank you!!
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May 29th, 2018, 02:21 AM   #2
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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.
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May 29th, 2018, 12:23 PM   #3
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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?
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June 18th, 2018, 06:06 AM   #4
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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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