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$a_1=1$, $a_2=3$, $a_{n+2}=a_{n+1}-2a_n-1$

How do you solve this? I only solve the sequence in the form $a_{n+2}=ba_{n+1}+ca_n$ before by writing it in $x^2-bx-c=0$ but for this I don't know how to. Please help

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  • $\begingroup$ it might also be fun/ helpful to find the generating series of it and develop a closed form using that $\endgroup$ – Control May 24 at 9:50
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The method is fairly similar, you suppose the solution is composed of 2 parts, namely the particular solution (which takes care of the constant and the homogenous solution which is the case you know how to solve. $$a_{n} = a^{p}_{n} + a^{h}_{n}$$ To solve for $a_{n}^{p}$ say it is some constant $c$ and plug it in to the reccurence relation. $$c = c - 2c - 1$$ $$2c = -1$$ $$c = -\frac{1}{2}$$ For the homogenous solution suppose the $-1$ isn't there, you get the following reccurence relation: $$a_{n+2} = a_{n+1} -2a_{n}$$ Which has the characteristic polynomial: $$x^2 - x + 2 = 0$$ $$x_{1, 2} = \frac{1 \pm \sqrt{1 - 8}}{2} = \frac{1 \pm i \sqrt{7}}{2}$$ Then our solution has the following form: $$a_{n} = \alpha(\frac{1 - i \sqrt{7}}{2})^n + \beta(\frac{1 + i \sqrt{7}}{2})^n - \frac{1}{2}$$ All that remains is to solve for $\alpha, \beta$ by evaluating it at $n=1, 2$.

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