(3-m)Sn+2man=m+3
(3-m)S(n-1)+2ma(n-1)=m+3
相减,且Sn-S(n-1)=an
所以(3-m)an+2man-2ma(n-1)=0
(3+m)an=2ma(n-1)
an/a(n-1)=2m/(3+m)是个常数
所以an是等比数列
q=2m/(3+m)
S1=a1
代入(3-m)Sn+2man=m+3
(3-m)a1+2ma1=m+3
a1=1
所以an=[2m/(3+m)]^(n-1)