1.已知单调递增的等比数列{an}满足a1+a2+a3=39,且a2+6是a1,a3的等差中项.

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1.已知单调递增的等比数列{an}满足a1+a2+a3=39,且a2+6是a1,a3的等差中项.
(1).求数列{an}的通项公式;
(2).设bn=3n/(an+1)(an+1+1).数列{bn}的前n项和为Sn,求证:Sn

1.
a1+a1*q+a1*q^2=39
a1(q^2+q+1)=39
a1=39/(q^2+q+1)(1)
2*(a1*q+6)=a1+a1*q^2,
a1(q^2-2q+1)=12
a1(q-1)^2=12
a1=12/(q-1)^2(2)
所以39/(q^2+q+1)=12/(q-1)^2
12(q^2+q+1)=39(q^2-2q+1)
12q^2+12q+12=39q^2-78q+39
3q^2-10q+3=0
(3q-1)(q-3)=0
q=1/3,(因为是增函数,所以舍去)
q=3
a1=12/(3-1)^2=3
所以an=a1+q^(n-1)
an=3+3(n-1)
2.
bn=3n/(an+1)(an+1+1)
=3n/[4+3^(n-1)](4+3^n)
=9n/(12+3^n)(4+3^n)
=9n*[1/(4+3^n)-1/(12+3^n)]/8
=9n/8*[1/(4+3^n)-1/(12+3^n)]
=9n/8(4+3^n)-9n/8(12+3^n)
Sn=9/56-9/120+18/8*13-18/8*21+ ---- +9n/8(4+3^n)-9n/8(12+3^n)

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