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Consider the recurrence

$$ a_n - 3a_{n-1} + 2a_{n-2} = 6 \cdot 2^n. $$

I tried to solve this as follows. First, I found the homogeneous solution: $$ a_n^{(h)} = r^2 - 3r + 2r \\ (r-2)(r-1) = 0 \\ r = 2,1 \\ a_n^{(h)} = A \cdot 2^n + B $$

While trying to solve for the particular solution (adding $n$ for linear independence), I get an answer which is a false statement: $$ a_n^{(p)} = Cn2^n \\ Cn2^n - 3[C(n-1)2^{n-1}] + 2[C(n-2)2^{n-2}] = 6\cdot 2^n \\ 4Cn - 3[2Cn-2] + 2[Cn-2] = 6 \cdot 2^2 \\ 4Cn - 6Cn + 6 + 2Cn - 4 = 24 \\ 0Cn+2 = 24 \\ 0Cn = 22 $$

I’m not sure where the issue lies. Am I performing the problem incorrectly, or does the problem have an issue?

Yuval Filmus
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critter1854
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1 Answers1

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Your calculation is an example of proof by contradiction. You assumed that $a_n = Cn2^n$, and reached a contradiction $0 = 22$. You can conclude that $a_n \neq Cn2^n$.

Your calculation shows that the general solution to your equation is $$ a_n = A + 2^n B + n2^n C. $$ For some reason you forgot about the first two terms.

Let $P(x) = \sum_{n=0}^\infty a_n x^n$. Setting $a_{-1} = a_{-2} = 0$, we have $$ (2x^2-3x+1) P(x) = \sum_{n=0}^\infty x^n (2a_{n-2} - 3 a_{n-1} + a_n) = \\ a_0 + (a_1 - 3a_0) x + \sum_{n=2}^\infty x^n (6 \cdot 2^n) = \\ a_0 + (a_1 - 3a_0) x + 24x^2 \sum_{m=0}^\infty (2x)^m = \\ a_0 + (a_1 - 3a_0) x + \frac{24x^2}{1-2x}. $$ Since $2x^2-3x+1 = (2x-1)(x-1)$, it follows that $$ P(x) = \frac{(1-2x)(a_0 + (a_1 - 3a_0) x) + 24x^2}{(1-2x)^2(1-x)} = \frac{a_0 + (a_1 - 5a_0)x + (24-2a_1+6a_0)x^2}{(1-2x)^2(1-x)}. $$ Routine calculation shows that $$ \frac{1}{(1-2x)^2(1-x)} = \sum_{n=0}^\infty (1+2^{n+1}n)x^n, $$ and so for $n \geq 2$ we get the formula $$ a_n = a_0 (1 + 2^{n+1}n) + (a_1 - 5a_0) (1 + 2^n(n-1)) + (24 - 2a_1 + 6a_0) (1 + 2^{n-1}(n-2)). $$ Opening this up, we get the solution $$ a_n = 24 - a_1 + 2a_0 - (24 - a_1 + a_0) \cdot 2^n + 12 \cdot n2^n. $$

Yuval Filmus
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