Proof verification: Is my induction proof correct? Can anyone proof read my induction proof for this particular task?
For $$ n \in \mathbb{N_0} $$
$a_n$ is defined as $$ a_n= (-1)^{(n+1)}+ 1 + (-3n) + 2n^2 $$
and $b_n$ as:
$$ b_n =\begin{cases}
          a_n \quad &\text{if} \, 0 ≤ n ≤ 2, \\
          b_{n-1}+b_{n-2}-b_{n-3}+8 \quad &\text{if} \, 3 ≤ n \\
     \end{cases}
$$
Prove or disprove by induction that $ a_n = b_n $
$$Induction Base: n=0,n=1,n=2,n=3$$
$$b_0 =a_0 =(−1)0+1 +1−3·0+2·02 =−1+1=0$$
$$b1 =a1 =(−1)1+1 +1−3·1+2·12 =1+1−3+2=1$$
$$b_2 =a_2 =(−1)2+1 +1−3·2+2·22 =−1+1−6+8=2 
$$
$$b_3 =b_2 +b_1 −b_0 +8=2+1−0+8=11 
$$
Induction Hypothesis: For all $k \in \mathbb {N_0}$ and k<n and n≥3 for some random but fixed $n \in \mathbb{N_0} $  a_k = b_k
Induction Step: $bn =b_{n−1} +b_{n−2} −b_{n−3} +8=(−1)^n +1−3(n−1)+2(n−1)$
$+(−1)^n−1 +1−3(n−2)+2(n−2)^2−(−1)^{n−2} +1−3(n−3)+2(n−3)^2+8$
$=(−1)(−1)(−1)^{n−2} +1−3(n−1)+2(n−1)^2 +(−1)^{n−1} +1−3(n−2)+2(n−2)^2$
$−(−1)^{n−2} −1+3(n−3)−2(n−3)^2 +8 = 1(−1)^{n−2} +1−3n+3+2n^2 −2^{n+1}+1(−1)^{n−1} −3n+6+2n^2 −4n+4$
$−(−1)^{n−2} +3n−9−2n^2 −6n+9+8= (11) =(−1)(−1)(−1)^{n−1} +9−3n+2n^2 −4n+2+2n^2 −8n+8−2n^2 +12n−18$
$=(−1)^{n+1} +1−3n+2n^2 =a_n $

Let me propose a bit cleaner alternative proof by strong induction.
You have already showed that these two sequences agree on first four terms $n=0,1, 2, 3.$ Then directly compute $a_{n-1}+a_{n-2}-a_{n-3}+8$ using the given formula, and make sure it is equal to $a_n$ for all $n\ge 3.$ Thus, both $(a_n), (b_n)$ satisfies the same recurrence relation. Fix an integer $k\ge 3,$ and assume the result: $a_n =b_n$ holds for all $3\le n\le k,$ then
\begin{align}
a_{k+1}-b_{k+1}
& = (a_{k}+a_{k-1}-a_{k-2}+8)-(b_{k}+b_{k-1}-b_{k-2}+8)\\
& = (a_k-b_k)+(a_{k-1}-b_{k-1})-(a_{k-2}-b_{k-2})\\
& = 0
\end{align}
proves the result for $n=k+1.$
