n=1时,S1=a1=6/7
a1=3S1/(1+3)=3a1/4,所以矛盾
n≥2时
Sn=(n+3)/3*an
an=Sn-Sn-1=(n+3)/3*an-(n-1+3)/3*an-1
(n+2)/3*an-1=n/3*an
an/an-1=(n+2)/n
an=(n+2)/n*an-1
a1=6/7
所以a2=4/2*6/7=12/7
a3=5/3*12/7=20/7
所以an=(n+1)(n+2)/7
所以a48=49*50/7=350
n=1时,S1=a1=6/7
a1=3S1/(1+3)=3a1/4,所以矛盾
n≥2时
Sn=(n+3)/3*an
an=Sn-Sn-1=(n+3)/3*an-(n-1+3)/3*an-1
(n+2)/3*an-1=n/3*an
an/an-1=(n+2)/n
an=(n+2)/n*an-1
a1=6/7
所以a2=4/2*6/7=12/7
a3=5/3*12/7=20/7
所以an=(n+1)(n+2)/7
所以a48=49*50/7=350