4Sn=(2n+3)an+1
4S(n-1)=[2(n-1)+3]a(n-1)+1
相减,Sn-S(n-1)=an
所以4an=(2n+3)an-(2n+1)a(n-1)
(2n-1)an=(2n+1)a(n-1)
an/a(n-1)=(2n+1)/(2n-1)
所以a(n-1)/a(n-2)=(2n-1)/(2n-3)
……
a3/a2=7/5
a2/a1=5/3
相乘并约分
an/a1=(2n+1)/3
a1=-3
an=-(2n+1)
1/a1a2+1/a2a3+```+1/an(an+1)
=-[1/3*5+1/5*7+……+1/(2n+1)(2n+3)]
=-(1/2)*[1/3-1/5+1/5-1/7+……+1/(2n+1)-1/(2n+3)]
=-(1/2)*[1/3-1/(2n+3)]
=-n/(6n+3)