数列{an}首项为1,且数列前n项和满足3mSn=(5m-2)Sn-1+3m(1)求证数列为等比数列（2）数列的公比为q=f(m)若数列bn的首项为4,且bn+1=f(1/bn)求数列bn的通项公式bn

Question

数列{an}首项为1,且数列前n项和满足3mSn=(5m-2)Sn-1+3m
(1)求证数列为等比数列
（2）数列的公比为q=f(m) 若数列bn的首项为4,且bn+1=f(1/bn)求数列bn 的通项公式bn

Answer 1

(1)3mSn=(5m-2)Sn-1+3m----1
3mSn-1=(5m-2)Sn-2+3m----2
1-2 3man=(5m-2)an-1
an=(5m-2/3m)an-1 公比q:5m-2/3m
(2)q=f(m)=5m-2/3m
bn+1=f(1/bn)=5(1/bn)-2/3(1/bn)=5/3-2/3*bn可化为(bn+1)-1=-2/3(bn-1)
b1+1=5 bn-1=5*(-2/3)^n-1
bn=5*(-2/3)^n-1+1