I have this task that gives me really a hard time:
For $$ n \in \mathbb{N_0} $$
$a_n$ is defined as $$ a_n= (-1)^{(n+1)}+ 1 + -3n + 2n^2 $$
and $b_n$ as: $$ b_n =\begin{cases} a_n \quad &\text{for} \, 0 ≤ n ≤ 2, \\ b_{n-1}+b_{n-2}-b_{n-3}+8 \quad &\text{for} \, 3 ≤ n \\ \end{cases} $$
Prove or disprove by induction that $ a_n = b_n $ for $n \in \mathbb{N} $
Can anyone help me with this? I have a hard time finding the right induction for this and its kinda important that i understand it. Would be great to see how to approach this properly. Thanks in advance!
Indeed you just need to verify that $\{a_n\}_{n\geq 0}$ satisfies the same recurrence relation: $$a_n=a_{n-1}+a_{n-2}-a_{n-3}+8.$$ And it is straightforward to find that's true. Let's see how to verify this point:
$$\begin{split} &a_{n-1}+a_{n-2}-a_{n-3}+8\\ ={}&(-1)^n+1-3(n-1)+2(n-1)^2+(-1)^{n-1}+1-3(n-2)+2(n-2)^2\\ &-(-1)^{n-2}-1+3(n-3)-2(n-3)^2+8\\ ={}&2n^2-3n+1-(-1)^{n-2}\\ ={}&a_n \end{split}$$ Suppose that $a_k=b_k$ for all $k\leq n$, then $$\begin{split} b_{n+1}&=b_{n}+b_{n-1}-b_{n-2}+8\\ &=a_n+a_{n-2}-a_{n-2}+8\\ &=a_{n+1}. \end{split}$$ According to the Strong Induction Principle, we know $a_n=b_n(\forall n\geq 0)$