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Non Homogenous Recurrence Relation: $$a_n=5a_{n-1}-6a_{n-2}+(13+6n)5^{n-2}$$

Most problems I've solved are of the form $a_n+a_{n-1}+a_{n-2}+F(N)$ I have never seen anything of this form its super weird.

Semiclassical
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    Welcome to MSE. Please type your questions rather than posting images. Images can't be browsed, and are not accessible to those using screen readers. If you need help formatting math on this site, here's a tutorial To begin with, surround all math expressions (including numbers,) with $ signs. Use ^ for exponents and _ for subscripts. $x_1^{2/3}$ shows up as $x_1^{2/3}$. – saulspatz Apr 21 '21 at 00:06
  • If you introduce $b_n:=a_n 5^{-n}$, then this recurrence relation becomes $$b_n =b_{n-1}-(6/25)b_{n-2}+(13+6n)/25$$ which eliminates the exponential scaling. (This helps to motivate the ansatz used by saulspatz below.) – Semiclassical Apr 21 '21 at 05:37

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We guess a solution of the form $a_n=(a+bn)5^n$ for some constants $a$ and $b$. Substituting into the recurrence and solving, I got $a=0, b=1$, so that a particular solution is $a_n=n5^n$.

EDIT

Substituting into the recurrence gives $$(an+b)5^n-5(a(n-1)+b)5^{n-1}+6(a(n-2)+b)5^{n-2}=(13+6n)5^{n-2}$$ Dividing through by $5^{n-2}$, $$ 25(an+b)-25(an-a+b)+6(an-2a+b)=13+6n\\ 25a+6an-12a+6b=13+6n$$ Equating coefficients, $$6a = 6\\ 13a+6b=13$$ and $$(a,b)=(1,0)$$

saulspatz
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