Question

高二数列

知an的前n项和sn满足2sn+an=1
求数列an的通项公式
设bn=2an+1/（1+an）（1+an+1） 数列bn的前n项和为tn，tn小于1/4

Answer 1

a[1] = s[1] = 1/3
a[n] = s[n]-s[n-1] (n>1)
3s[n] - s[n-1] =1
(3^n)s[n]-(3^(n-1))s[n-1] = 3^(n-1)
M[n] = 3^ns[n]
M[1] = 3s[1] =1
M[n] - M[n-1] = 3^(n-1)
M[n] -M[1] = 3^(n-1)+...+3^(2-1)
M[n] = (3^n-1)/(3-1) = (3^n-1)/2
S[n] = M[n]/3^n=(1-3^(-n))/2
a[n] =1 - 2S[n] = 3^(-n)
b[n] = 2*3(-n-1)/(1+3^(-n))(1+3^(-n-1)) = 2*3^n/(1+3^n)(1+3^(n+1))
= 1/(1+3^n)-1/(1+3^(n+1))
t[n] = sum(b[n]) = 1/4-1/(1+(3^(n+1)))<1/4