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I have been trying to solve the following recurrence relation: $a_{n+2}=4a_{n+1}-4a_n+n^2-5n+2$. I already figured out that the general solution should be $c_12^n+c_2n2^n$ because the root 2 has multiplicity 2. But I'm having hard time finding the particular solution to this recurrence relation. Any help will be appreciated.

Ben
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    What is "advanced operator A"? Also, have you tried just writing out the first few terms and simplifying? – Milo Brandt Dec 02 '19 at 03:07
  • It's something I learned in my combinatorial math course. Define Af(n)=f(n+1) in recurrence relations. – Ben Dec 02 '19 at 03:19
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    Since you didn't give a starting value, I took $y_0=1$ and made a spreadsheet. I found some numbers that are old friends. The square root might make it easier to make a recurrence for $y_i^2$ – Ross Millikan Dec 02 '19 at 03:38
  • Thank you! I solved it using ansatz. :) – Ben Dec 02 '19 at 04:52

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To get rid of the value of $y_0$, define $u_n=\frac {y_n}{y_0}$ and later $v_n=u_n^2$ to face $$v_{n+1}=\frac{n+3}{n+1}\,v_n\qquad \text{with} \qquad v_0=1$$ Comupute the very first terms and notice that they are along a quadratic in $n$. So, let $v_n=a+b n+c n^2$ and identify the coefficients.

Claude Leibovici
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