行列式按第1行展开,再展开一次即得
A2n = andnA2(n-1)+(-1)^(1+2n)bn(-1)^(2n-1+1)cn
= (andn-bncn)A2(n-1)
递归得
A2n = (andn-bncn)A2(n-1)
= (andn-bncn)(an-1dn-1 - bn-1cn-1)A2(n-2)
= ...
= ∏(aidi-bici)