Question

高中一道数学题

数列an的前n项和为Sn,满足关系;1-sn+1=an-a(n+1)，n属于自然数，设bn=(（1/log2（a(2n+3))-2/log2(a(2n+1)))an,数列Bn的前n项和为Tn，求Tn
Sn,An,Tn,n为第几项

Answer 1

1-sn+1=an-a(n+1) 1-sn+1=an-S(n+1)+Sn Sn=1-an
an=(1-an)-[1-a(n-1)] 2an=a(n-1)
1-(a1+a2)=a1-a2 a1=1/2 an=(1/2)*(1/2)^(n-1)=2^(-n) Sn=1-2^(-n)

bn=(（1/log2（a(2n+3))-2/log2(a(2n+1)))an
=1/log2[2^(-(2n+3))]-2/log(2^(-(2n+1))]
=-1/(2n+3)+1/(2n+1)
Tn=1/3-1/5+1/5-1/7+……+1/(2n+1)-1/(2n+3)=1/3-1/(2n+3)=2n/[3*(2n+3)]