當p=1時
an-a(n-1)=q
an=a1+q(n-1)
當p≠1時
an=pa(n-1)+q
an=pa(n-1)+q(1-p)/(1-p)
an=pa(n-1)+q/(1-p)-pq/(1-p)
an-q/(1-p)=pa(n-1)-pq/(1-p)
an-q/(1-p)=p[a(n-1)-q/(1-p)]
[an-q/(1-p)]/[a(n-1)-q/(1-p)]=p
故數(shù)列{an-q/(1-p)}是公比為p的等比數(shù)列
則an-q/(1-p)=[a1-q/(1-p)]p^(n-1)
an=[a1-q/(1-p)]p^(n-1)+q/(1-p)
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