已知数列an的前n项和公式为Sn=kq^n-k,求证数列an为等比数列

Question

已知数列an的前n项和公式为Sn=kq^n-k,求证数列an为等比数列

Answer 1

∵Sn=kq^n-k
∴S(n+1)=kq^(n+1)-k
∴a(n+1)=S(n+1)-Sn
=[kq^(n+1)-k]-(kq^n-k)
=k[q^(n+1)-q^n]
=k[(q-1)q^n
a(n+1)/an=k[(q-1)q^n/[k(q-1)q^(n-1)=q
∴数列{an}为等比数列