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Let

$$a_n - 3a_{n-1} + 2a_{n-2} = 2n - 1$$ $$a_0 = a_1 = 0$$

Be a recurrence relation. To solve it, we need to find homogenous and particular solution. I found out that the homogenous solution is $b_1 + b_2 \cdot 2^n$ for some constants $b_1, b_2$. To find the particular solution, I say that $a_n ^p = cn + d$ for some constants $c, d$. But when we put it into the recurrence, we get that

$$cn+d - 3(cn - c + d) + 2(cn - 2c + d) = 2n-1$$ $$ (c-3c+2c)n + (d+3c-3d-4c+2d) = 2n-1 $$ $$ -c = 2n-1 $$

and therefore we cant find $d$, so probably guessing that $a_n ^p = cn + d$ is wrong. How do i find the particular solution then? Because I thought that the particular solution is $cn+d$ when $d(n) = \alpha n + \beta$ ( where $d(n)$ is the non homogenous part of the recurrence )... Thanks to everyone helping!

Kevinlove
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    Because 1 is a root of the characteristic equation, you will need to and use a quadratic (one degree higher) instead of a linear term. – Calvin Lin Mar 03 '24 at 15:18
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    If you sum together the first $n$ terms of the relation, you end up with $$a_n-2a_{n-1}=n^2-1$$ Perhaps this will be easier to solve. – Rhys Hughes Mar 03 '24 at 16:22

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Let $$\alpha(x):= \sum_{n=0}^\infty a_n x^n$$

We have $$ (1-3x+2x^2) \alpha(x)= \sum_{n=2}^\infty \left(a_n-3a_{n-1}+2a_{n-2} \right) x^n=\sum_{n=2}^\infty (2n-1) x^n=\frac{(3-x)x^2}{(x-1)^2} $$ (see here ). So $$ \alpha(x)=\frac{(3-x)x^2}{(x-1)^2(1-3x+2x^2)} $$

Its Taylor series is given by $$ \alpha(x)=\sum_{n=0}^\infty \left(-n(n+4)+5 (2^n-1) \right) x^n $$ (see here) and so we have $$ a_n=\left(-n(n+4)+5 (2^n-1) \right) $$

Marco
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