前n项和为sn,a1=1,an+1=(n+2/n)sn,求通项公式.如题n+1为角标

Question

前n项和为sn,a1=1,an+1=(n+2/n)sn,求通项公式.
如题
n+1为角标

Answer 1

1、A(n+1)=(n+2)sn/n=S(n+1)-Sn
即nS(n+1)-nSn=(n+2)Sn
nS(n+1)=(n+2)Sn+nSn
nS(n+1)=(2n+2)Sn
S(n+1)/(n+1)=2Sn/n
即S[(n+1)/(n+1)]/[Sn/n]=2
S1/1=A1=1
所以Sn/n是以2为公比1为首项的等比数列
2、由1有Sn/n是以2为公比1为首项的等比数列
所以Sn/n的通项公式是Sn/n=1*2^(n-1)
即Sn=n2^(n-1)
那么S(n+1)=(n+1)2^n,S(n-1)=(n-1)2^(n-2)
An=Sn-S(n-1)
=n2^(n-1)-(n-1)2^(n-2)
=n*2*2^(n-2)-(n-1)2^(n-2)
=[2n-(n-1)]*2^(n-2)
=(n+1)2^(n-2)
=(n+1)*2^n/2^2
=(n+1)2^n/4