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In answering a question myself I have ended up with a summation of the form: $$a_n=\frac{n(n+3)}{2}a_{n-1}+\frac{n(n+1)}{2}$$ I am unsure of how to solve this type of summation such that I have it in the form: $$a_n=f(a_0,n)$$ I know that problems of this form can be solved as I have seen it in the Fibonacci sequences and other cases but I am not sure how I go about doing this, I will link the original question here. Thanks!

Evidence of my attempt so far and how I got in this form can be found at the original question

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I've found a formula here which talks about how to solve first order non-homogeneous recurrence relations, which states for a series in the form: $$a_{n+1}-f_na_n=g_n$$ we can say: $$a_n=\left(\prod_{k=0}^{n-1}f_k\right)\left(A_0+\sum_{m=0}^{n-1}\frac{g_m}{\prod_{k=0}^mf_k}\right)$$ So I will try to use this

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  • $\begingroup$ Do you know about generating functions? $\endgroup$ – kccu May 6 '19 at 22:55
  • $\begingroup$ Not really no, although by searching it, it seems it could be used to solve problems of this form $\endgroup$ – Henry Lee May 6 '19 at 22:56
  • $\begingroup$ Yes, that would be my approach. $\endgroup$ – kccu May 6 '19 at 23:00
  • $\begingroup$ Is there a way you could explain how to apply it when it is in this form? I can't find an example where $a_{n-1}$ is multiplied by a function of $n$ $\endgroup$ – Henry Lee May 6 '19 at 23:02
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Let $A(x)=\sum_{n=0}^\infty a_nx^n$. Your recurrence can be re-written as: $$a_{n+1}=\frac{(n+1)(n+4)}{2}a_n + \frac{(n+1)(n+2)}{2} = \frac{(n+1)(n+2)}{2}a_n + (n+1)a_n + \frac{(n+1)(n+2)}{2}.$$ Now we multiply through by $x^n$ and sum from $n=0$ to $\infty$: \begin{align*} \sum_{n=0}^\infty a_{n+1}x^n&= \frac{1}{2}\sum_{n=0}^\infty (n+1)(n+2)a_nx^n + \sum_{n=0}^\infty (n+1)a_nx^n + \frac{1}{2}\sum_{n=0}^\infty (n+1)(n+2)x^n\\ \frac{1}{x}\sum_{n=0}^\infty a_{n+1}x^{n+1}&= \frac{1}{2}\sum_{n=0}^\infty (n+1)(n+2)a_nx^n + \sum_{n=0}^\infty (n+1)a_nx^n + \frac{1}{2}\sum_{n=0}^\infty (n+1)(n+2)x^n\\ \frac{1}{x}(A(x)-a_0)&= \frac{1}{2} \frac{d^2}{dx^2}A(x) + \frac{d}{dx}A(x) + \frac{1}{2}\frac{d^2}{dx^2} \Bigg(\underbrace{\sum_{n=0}^\infty x^n}_{\frac{1}{1-x}}\Bigg)\\ \frac{1}{x}(A(x)-a_0)&= \frac{1}{2} A''(x) + A'(x) + \frac{1}{(1-x)^3}. \end{align*} So now you would need to solve this second-order non-homogeneous differential equation.

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  • $\begingroup$ That is great, thank you! $\endgroup$ – Henry Lee May 6 '19 at 23:40

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