How do I prove that $\det A_{n+2} = a \det A_{n+1} + b \det A_n$ for matrix $A$? 
I have calculated:
$\det A_1=2$, $\det A_2=3$, $\det A_3=4$, so I was putting some numbers in $\det A_{n+2} = a \det A_{n+1} + b\det A_n$ like $n=1$, $n=2$ ($\det$ $n\times n$ matrix) and get that $a=2$, $b=-1$, 
but how that can help me to prove that $\det A_{n+2} = a\det A_{n+1} + b\det A_n$?
And the end how do I prove that $\det A_n = n+1$?
 A: it's easy indeed. to calculate $\det(A_{n+2})$ expand determinant from first column and you get this:
$$\det(A_{n+2})=2\det(A_{n+1})-\det(B)$$
where $\det(B)$ is $\det(A_{n+2})$ with the first column and second row removed. now again expand $\det(B)$ in first row to get $\det(B)=\det(A_{n})$ and proof is complete.
to solve the second part $\det(A_{n})=n+1$, can be done by induction.
A: Use induction on $n$. For $n=1$, the result is trivial. Now,
suppose that there is an $N\in\mathbb{N}$ such that $\det A_n=n+1$ for all 
$n\le N$. Then for the case $N+1$, we have
\begin{align}
\det A_{N+1}
&=2\det A_N-\det
\begin{pmatrix}
2&1&0&0&&&&&\\
1&2&1&0&&&&&\\
0&1&2&1&&&&&\\
&&&&\ddots\\
&&&&&1&2&1&0\\
&&&&&0&1&2&0\\
&&&&&0&0&1&1
\end{pmatrix}_{N\times N}\\
&=2(N+1)-\det
\begin{pmatrix}
2&1&0&0&&&\\
1&2&1&0&&&\\
0&1&2&1&&&\\
&&&&\ddots\\
&&&&&1&2&1\\
&&&&&0&1&2
\end{pmatrix}_{(N-1)\times(N-1)}\\
&\quad+\det
\begin{pmatrix}
2&1&0&0&&&\\
1&2&1&0&&&\\
0&1&2&1&&&\\
&&&&\ddots\\
&&&&&1&2&0\\
&&&&&0&1&0
\end{pmatrix}_{(N-1)\times(N-1)}\\
&=2N+2-[(N-1)+1]+0\\
&=N+2.
\end{align}
This completes the proof for the second question.
