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.

来源：学生作业帮编辑：拍题作业网作业帮分类：数学作业时间：2024/05/06 01:21:48

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)

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. 已知等比数列{an}满足2a1+a3=3a2,且a3+2是a2,a4的等差中项已知等比数列｛an｝是递增数列,且满足a1+a2+a3=39,a2+6是a1和a3的等差中项求｛an｝的通项公式已知单调递增的等比数列 an满足,a2+a3+a4=28,且a3+2是a2,a4a的等差中项已知单调递增的等比数列an满足：a2+a3+a4=28,且a3+2是a2,a4等差中项已知等比数列(an)满足2a1+a3=3a2且a3+2是a2,a4的等差中项求数列(an)的通项公式? 已知等比数列{an}满足2a1＋a3＝3a2.且a3+2是a2.a4的等差中项.求数列已知递增的等比数列{an}满足a2+a3+a4=28,且a3+2是a2,a4的等差中项,求数列{an}的通项公式已知递增的等比数列{an}满足a2+a3+a4=28,且a3+2.是a2.a4的等差中项,求{an}的通项公式已知递增的等比数列{an}满足a2+a3+a4=28，且a3+2是a2，a4的等差中项．已知递增的等比数列{an}满足a2+a3+a4+28,且a3+2是a2,a4的等差中项,求数列{an}的通项公式已知单调递增的等比数列an,a2+a3+a4=28,a3+2是a2和a4的等差中项,求an的通项公式.