Given $\frac{b_{n}}{b}= trace S^{k} A= \displaystyle \sum_{j=0}^k=\displaystyle \sum_{i=0}^{min ({k-j, j)}} {k-j \choose i}{j \choose j-i}a^{k-j-i}b^{i}c^id^{j-i}$ where $n=k+1, s=i \ $ and $ \ m=j+i $. Let $A^n=A^{n-1}A= \ M= \left( {\begin{array}{cc} a_{n-1} & b_{n-1} \\ c_{n-1} & d_{n-1} \\ \end{array} } \right)\left( {\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right)$.
And I get the recursive equations: $$\begin{cases}a_n=aa_{n-1}+cb_{n-1}\\b_n=ba_{n-1}+db_{n-1}\\c_n=ac_{n-1}+cd_{n-1}\\d_n=bc_{n-1}+dd_{n-1}\end{cases}$$
Show that $$b_n = \displaystyle \sum_{s=0}^{[\frac{n-1}{2}]} \displaystyle \sum_{m=0}^{{n-2s-1}} {n-s-m-1 \choose s}{m+s \choose m}a^{n-2s-m-1}b^{s+1}c^s d^{m}$$. However, I don't know what is $b_{n-1}$ and $a_{n-1}.$ I just knew that it uses recurrence equation. How, will I prove this?