Let $K_n \in \mathbb{R}^{n \times n}$, specifically for general $n \in \mathbb{N}$ $$ (K_n)_{ij} = \begin{cases} 2&, i=j\\ -1&, \|i-j\| = 1\\ 0 &, \text{otherwise} \end{cases} $$ where $1 \leq i,j \leq n$. For example for $n=3$ this results in $$ K_3 = \begin {pmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end {pmatrix}. $$ Prove that det$(K_n) = n+1$.
My first approach is to use a proof by induction. By calculating the determinant for the first five matrices using laplace's formula I found that for $n \geq 4$ the determinant seems to satisfy $$ \text{det}(K_n) = 2 \text{det}(K_{n-1}) - \text{det}(K_{n-2}). $$
Now it has been quite a while since i last used induction and I am having trouble forumlating the induction step. Naively i would argue that for $n+1$ we get $$ \text{det}(K_{n+1}) = 2 \text{det}(K_n) - \text{det}(K_{n-1}) =2 (n+1) - n = n+2 $$ which I think concludes the proof, but I'm unsure, since I used the recursive determinant formula for $n+1$ as well as my induction hypothesis. Would i need to proof the determinant formula seperately? If so, how would one go about it?