Question

已知数列An前n项和Sn=2An-3·2^n+4,求证数列｛An/(2^n)｝为等差数列并求数列An的通项公式

另(2)：设Tn为数列{Sn-4}的前n项和，求Tn

Answer 1

(1)
s1=a1=2a1-3*2+4，得a1=2
sn=2an-3*2^n+4
s(n-1)=2a(n-1)-3*2^(n-1)+4
sn-s(n-1)=2an-3*2^n+4-2a(n-1)+3*2^(n-1)-4=an
得an-3*2^n=2a(n-1)-3*2^(n-1)
(an/2^n)-3=[a(n-1)/2^(n-1)]-(3/2)
得an/2^n=[a(n-1)/2^(n-1)]+(3/2)，a1/2=1
于是数列{an/2^n}是以1为首项，(3/2)为公差的等差数列
an/2^n=(3n-1)/2
an=(3n-1)*2^(n-1)
(2)
sn=2an-3*2^n+4=(3n-4)*2^n+4
sn-4=(3n-4)*2^n
Tn=-1*2+2*2²+5*2³+8*2^4+……+(3n-4)*2^n
2Tn=-1*2²+2*2³+5*2^4+8*2^5+……+(3n-4)*2^(n+1)
Tn-2Tn=-1*2+3(2²+2³+2^4+……+2^n)-(3n-4)*2^(n+1)【错位相减】
-Tn=-(3n-7)*2^(n+1)-14
Tn=（3n-7)*2^(n+1)+14

Answer 2

(1)
Sn = 2an- 3.2^n +4
n=1 , a1= 2
an = Sn-S(n-1)
= 2an - 2a(n-1) - 3.2^(n-1)
an = 2a(n-1) + 3.2^(n-1)
an/2^n - a(n-1)/2^(n-1) = 3/2
{ an/2^n }是等差数列, d=3/2
an/2^n - a1/2 = 3(n-1)/2
an/2^n = (3n-1)/2
an = (3n-1).2^(n-1)
(2)
Sn -4= 2an - 3.2^n
= (3n-1).2^n - 3.2^n
= (3n-4).2^n
= 3(n.2^n) - 4.2^n
Tn = 3[∑(i:1->n) (i.2^i) ] - 8(2^n-1)
let
S = 1.2 + 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)
Tn = 3[∑(i:1->n) (i.2^i) ] - 8(2^n-1)
= 3[n.2^(n+1) - 2( 2^n-1)] - 8(2^n-1)
=14 + ( 6n -14).2^n