Question

已知数列｛an｝的前n项和为Sn，Sn=2an-2。设b n=an•log2^an+1，求数列｛b

已知数列｛an｝的前n项和为Sn，Sn=2an-2。设b n=an•log2^an+1，求数列｛bn｝的前n项和Tn

Answer 1

本题考查顷搏册雀宏了利用“银档n=1，a1=S1；n≥2，an=Sn-Sn-1”求an、“错位相减法”、等比数列的通项公式及其前n项和公式等基础知识与基本技能方法

Answer 2

Sn=2an-2
n=1, a1=2
an =Sn -S(n-1)
=2an -2a(n-1)
an = 2a(n-1)
=2^(n-1).a1
=2^n
let
S =1.2^1+2.2^2+...+n.2^n (1)
2S = 1.2^2+2.2^3+...+n.2^(n+1) (2)
(2)-(1)
S = n.2^(n+1) -(2+2^2+...+2^n)
= n.2^(n+1) -2(2^n -1)

bn=an•log2^a(n+1)
= (n+1).2^n
= n.2^n + 2^n
Tn=b1+b2+...+bn
=S +2(2^n-1)
=n.2^(n+1)