设{an}前n项和Sn=2an-4(n属于自然数）,数列{bn}满足bn+1=an+2bn,b1=2

题目

设{an}前n项和Sn=2an-4(n属于自然数）,数列{bn}满足bn+1=an+2bn,b1=2
注题中n和n+1均为下标
（1）求an
(2)证明{bn/2^n}为等差数列
（3)求{bn}的前n项和Tn

答案

(1)由Sn=2an-4
则:S(n-1)=2a(n-1)-4
两式相减,则:
an=2an-2a(n-1)
2a(n-1)=an
an/a(n-1)=2
又a2/a1=8/4=2
则:an=a1*2^(n-1)
=4*2^(n-1)
=2^(n+1)
(2)由b(n+1)=an+2bn
则:b(n+1)=2bn+2^(n+1)
两边同时除以2^(n+1)
b(n+1)/[2^(n+1)]=2bn/[2^(n+1)]+1
b(n+1)/[2^(n+1)]=bn/[2^n]+1
{b(n+1)/[2^(n+1)]}-{bn/[2^n]}=1
则{bn/2^n}为等差数列
(3)由于{bn/2^n}为等差数列
则:bn/2^n=b1/2^1+(n-1)*1
=1+n-1
=n
则:bn=n*2^n
则:Tn=2+2×2^2+3×2^3+4×2^4+……+n×2^n
2Tn=1×2^2+2×2^3+3×2^4+……+(n-1)×2^n+n×2^(n+1)
∴Tn=n×2^(n+1)-2-(2^2+2^3+2^4+……+2^n）
Tn=n×2^(n+1)-2^(n+1)+2