the question is : Let Dn = $[a_{ij}]$n × n be a (n × n) determinant with the following conditions: $$a_{ij} = 4 ;i=j$$ $$a_{ij} = 2 ;|i-j|=1$$ $$a_{ij} = 0 ; otherwise$$ then we have to prove that : $$ D_n = 4D_{n-1} - 4D_{n-2}$$ my approach:
I found $D_1 =4 ,D_2=12, D_3=32$ but I don't know how to prove this as higher order determinants are take much more time to get calculated and I couldn't find a pattern .Thanks for all help.
\begin{align} D_n &=\overbrace{\begin{vmatrix} 4&2&0&\cdots&0&0&0\\ 2&4&2&\cdots&0&0&0\\ 0&2&4&\cdots&0&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ 0&0&0&\cdots&4&2&0\\ 0&0&0&\cdots&2&4&2\\ 0&0&0&\cdots&0&2&4\\ \end{vmatrix}}^{n\text{ columns}}\\ &= 4\cdot\overbrace{\begin{vmatrix} 4&2&0&\cdots&0&0&0\\ 2&4&2&\cdots&0&0&0\\ 0&2&4&\cdots&0&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ 0&0&0&\cdots&4&2&0\\ 0&0&0&\cdots&2&4&2\\ 0&0&0&\cdots&0&2&4\\ \end{vmatrix}}^{(n-1)\text{ columns}} -2\cdot\overbrace{\begin{vmatrix} 2&2&0&\cdots&0&0&0\\ 0&4&2&\cdots&0&0&0\\ 0&2&4&\cdots&0&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ 0&0&0&\cdots&4&2&0\\ 0&0&0&\cdots&2&4&2\\ 0&0&0&\cdots&0&2&4\\ \end{vmatrix}}^{(n-1)\text{ columns}}\quad\text{(expanding first row)}\\ &= 4\cdot\overbrace{\begin{vmatrix} 4&2&0&\cdots&0&0&0\\ 2&4&2&\cdots&0&0&0\\ 0&2&4&\cdots&0&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ 0&0&0&\cdots&4&2&0\\ 0&0&0&\cdots&2&4&2\\ 0&0&0&\cdots&0&2&4\\ \end{vmatrix}}^{(n-1)\text{ columns}} -4\cdot\overbrace{\begin{vmatrix} 4&2&0&\cdots&0&0&0\\ 2&4&2&\cdots&0&0&0\\ 0&2&4&\cdots&0&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ 0&0&0&\cdots&4&2&0\\ 0&0&0&\cdots&2&4&2\\ 0&0&0&\cdots&0&2&4\\ \end{vmatrix}}^{(n-2)\text{ columns}}\quad\text{(expanding first column)}\\ &=4D_{n-1}-4D_{n-2}\\ \end{align}