已知数列an中,a1=3,a2=5,其前n项和Sn满足Sn+Sn-2=2Sn-1+2^n-1(n>=3).令bn=2^n-1/an*an+1;Tn是数列{bn}的前n项和,证明Tn

Question

已知数列an中,a1=3,a2=5,其前n项和Sn满足Sn+Sn-2=2Sn-1+2^n-1(n>=3).
令bn=2^n-1/an*an+1;Tn是数列{bn}的前n项和,证明Tn

Answer 1

Sn+Sn-2=2Sn-1+2^n-1
移项可得
Sn-Sn-1=Sn-1 - Sn-2 +2^(n-1)
即
an=an-1 + 2^(n-1) (n>=3)
所以数列an的通项为
an= a1+ 2^1+2^2+...+2^(n-1)
= a1+ 2^n-2 (n>=3)
此通项可与a1=3,a2=5合并,得到
an=2^n+1
则有
bn=2^n-1/(2^n+1)(2^(n+1)+1)
=0.5*[1/(2^n+1)-1/(2^(n+1)+1)]
所以Tn=0.5*[1/3-1/5+1/5-1/9+...+1/(2^n+1)-1/(2^(n+1)+1)] ---注意中间项都正负抵消
=0.5*[1/3-1/(2^(n+1)+1)]