Question

设等比数列an的前n项和为Sn，已知a1＝2，且4S1，3S2，2S3成等差数列。(1)求数列an

设等比数列an的前n项和为Sn，已知a1＝2，且4S1，3S2，2S3成等差数列。(1)求数列an的通项公式 (2)设bn＝|2n－5|an，求数列bn的前n项和Tn

Answer 1

(1)
an=a1.q^(n-1) = 2.q^(n-1)
Sn = a1+a2+...+an

4S1，3S2，2S3成等差数列
4S1+2S3 = 6S2
8+2(2+q+q^2)= 6(2+q)
4+(2+q+q^2)= 3(2+q)
q^2-2q=0
q=2
an=2^n

(2)
let
S = 1.2^1 +2.2^2+.....+n.2^n (1)
2S = 1.2^1 +2.2^2+.....+n.2^(n+1) (2)
(2)-(1)
S = n.2^(n+1) -(2^1+2^2+...+2^n)
= n.2^(n+1) -2(2^n -1)
= 2 + (2n-2).2^n

bn=|2n-5|an
2n-5 >0
n> 5/2
ie
bn = -(2n-5)an ; n= 1,2
=(2n-5)an ; n=3,4,....

for n<=2
bn = -(2n-5). 2^n
= -2n.2^n + 5.2^n
Tn = b1+b2+...+bn
= -2S + 10(2^n-1)
=-2(2 + (2n-2).2^n) + 10(2^n-1)
= -14 -(2n-12).2^n

b1=3
b2 =4

n>=3
Tn
=(b1+b2) + (b3+b4+...+b5)
=(3+4) +(2S - 5.2^n) -(3+4)
=2(2 + (2n-2).2^n) - 10(2^n-1)
=14 +(2n-12).2^n