数列an.a1=1 3tSn-(2t+3)S(n-1)=3t
(t>0 ,n=2,3,.....)
求证：an是等比数列
公比是f(t)，做数列bn,使b1=1,bn=f[1/(bn-1)]求bn
求b1b2-b2b3+b3b4-....-b2nb(2n+1)的值

S2=a2+1
3t(a2+1)-(2t+3)=3t
===>a2=(2t+3)/3t
a2/a1=(2t+3)/3t ...................(1)
3tSn-(2t+3)S(n-1)=3t ........(2)
3tS(n-1)-(2t+3)S(n-2)=3t.......(3)
(2)-(3)
3tan-(2t+3)a(n-1)=0
an/a(n-1)==(2t+3)/3t ,(n=2,3.....)......(4)
(1),(4)==>an是a1=1 ,q=(2t+3)/3t 的等比数列

f(t)=(2t+3)/3t =2/3+1/t
bn=f[1/(bn-1)] =2/3 +b(n-1)
bn是b1=1 ,d=2/3 的等差数列
bn=1+(2/3)(n-1) =(2n+1)/3

b1b2-b2b3+b3b4-....-b2nb(2n+1)
=b2(b1-b3)+b4(b3-b5)+....+b2n[b(2n-1)-b(2n+1)]
=(-4/3)[b2+b4+.....+b2n] ................(5)
bn是b1=1 ,d=2/3 的等差数列
==>b2n是b2=5/3 ,d=4/3 的等差数列
b2n=(4n+1)/3
b2+b4+.....+b2n =(1/2)n[(5/30+(4n+1)/3] ........(6)
(6)--->(5)
==>
b1b2-b2b3+b3b4-....-b2nb(2n+1)
=-(4/9)n(2n+3)