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

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