how to solve this recursive relation

Question

please help me solve this recursive relation :

$$a_n-2a_{n-1}+a_{n-2} = n-2,$$ $$ a_0 = 1, a_1 = 2, n\geq 2$$ looks like non homogenous function but I can't reach to answer.

Answer 1

Two things to recognize: $$ a_n - 2a_{n-1} + a_{n-2} = \Delta^2 a $$ (the second difference operator), and the equation is linear in $a$ so that you can add separate answers for $n$ and $-2$ on the right.

Also, summation behaves like integration, using "rising powers" instead of just powers (see Concrete Mathematics by Knuth et al). Then:

$$ \Delta^2 a = n \Longrightarrow a_n = \frac{n^{\overline 3} }{6} = \frac{n(n+1)(n+2)}{6} $$ $$ \Delta^2 a = -2 \Longrightarrow a_n = -2\frac{n^{\overline 2} }{6} = -n(n+1) $$ Finally, $\Delta^2 (\alpha n + \beta) = 0$.

So the solution will be of the form $$ a_n = \frac{n^3-3n^2-4n}{6} + \alpha n + \beta$$

(The $4n$ term could be absorbed into $\alpha n$.)

Use $a_0 = 1$ to find $\beta = 1$, and then use $a_1 = 2$ to find $\alpha$. The answer is

$$ a_n = \frac{n^3-3n^2+8n+6}{6} $$

Answer 2

Let $y_n=a_n-a_{n-1}$, then $y_n-y_{n-1}=n-2$ and $y_1=1$. You can get $$ \sum_{k=2}^n y_k-y_{k-1}=\sum_{k=2}^n (k-2)=\frac{(n-2)(n-3)}{2}=y_n-y_1. $$ Thus, you can solve $y_n$, use the same trick to solve $a_n$.