题目

等差数列{an},a1=1,前n项和Sn,S2n/Sn=4
（1）求数列｛an｝的通项共识和Sn
（2）记bn=an*2^（n-1）求｛bn｝的前N项和Tn

提问时间：2021-02-28

答案

S2n=2n+n*(2n-1)d Sn=n+n(n-1)d/24Sn=4n+2(n^2-n)dS2n/Sn=4 S2n=4Sn4n+2d(n^2-n)=2n+(2n^2-n)d整理,得dn=2nd=2S2n=2n+n*(2n-1)d Sn=n+n(n-1)d/24Sn=4n+2(n^2-n)dS2n/Sn=4 S2n=4Sn4n+2d(n^2-n)=2n+(2n^2-n)d整理,得dn...