Question

已知｛an｝是公差不为零的等差数列，a1=1，且a1，a3，a9成等比数列。

数列通项为An=n，求数列「An*2的An次方」的前N项和Sn

Answer 1

an=a1+(n-1)d
a1=1
a1，a3，a9成等比数列
a1.a9 = (a3)^2
1+8d = (1+2d)^2
4d^2-4d=0
d=1
an =n

bn = an . 2^an
= n.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)
= 2+ (2n-2).2^n
Sn = b1+b2+...+bn = S =2+ (2n-2).2^n

Answer 2

设数列为bn
bn=2^(an）=2^n为首项为2 公比为2的等比数列所以Sn=b1(1-q^n)/(1-q)=2（1-2^n)/(1-2)=2^（n+1）-2