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.
问题描述：
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
答
1.a1+a1*q+a1*q^2=39 a1(q^2+q+1)=39a1=39/(q^2+q+1)(1)2*(a1*q+6)=a1+a1*q^2,a1(q^2-2q+1)=12a1(q-1)^2=12 a1=12/(q-1)^2(2)所以39/(q^2+q+1)=12/(q-1)^212(q^2+q+1)=39(q^2-2q+1)12q^2+12q+12=39q^2-78q+393q^2-10...