Question

已知数列{an}的前n项和Sn=-an-1/2^n-1+2(n为整数）（求详细、准确解。主要是第二题）

(1)令bn=2^n×an,求证数列{bn}是等差数列,并求数列{an}的通项公式
(2）令Cn=(n+1)|n*an,求Tn=c1+c2+...+cn的值

Answer 1

(1)
Sn=-an-1/2^n-1+2
Sn+1=-an+1-1/2^n+2
Sn+1-Sn=an+1=-an+1+an+1/2^n
2an+1=an+1/2^n
2^(n+1)an+1=2^nan+1
bn+1=bn-2 数列{bn}是等差数列
an=n/2^n
(2)
Cn=(n+1)/2^n
Tn= 2/2+3/4+4/8.............n/2^n-1+(n+1)/2^n
2Tn=2+3/2+4/4.....................+(n+1)/2^n-1
2Tn-Tn=2+1/2+1/4.....................+1/2^n-1-(n+1)/2^n
=2+(1-1/2^n-1)-(n+1)/2^n
=3-1/2^n-1-(n+1)/2^n
用错位相减，希望满意