Question

17题求解！！！急

Answer 1

1) a_n = (1/2)^{n-1}
because
S_n + a_n = S_{n-1} + a_{n-1} ==> S_n - S_{n-1} + a_n = a_{n-1} ==> a_n = 1/2 a_{n-1}
a_1 = S_1 = 1

2) log_2 (a_{n+1}) . log_2(a_{n+1})
=( log_2 ( 2^{-n} ) )^2
= n^2
==> 1/( n b_n) = n
==> T_n = (n+1) n /2

Answer 2

∵S(n)+a(n)=2.......(1)
∴S(1)+a(1)=2
∴2a(1)=2
a(1)=1
又S(n-1)+a(n-1)=2......(2)
(1)-(2)，得
2a(n)-a(n-1)=0
即a(n)=a(n-1)/2
∴{a(n)}是首项为a1=1, 公比为q=1/2的等比数列
a(n)=a(1)*q^(n-1)=1/2^(n-1)

将a(n+1)=1/2^n, a(n+2)=1/2^(n+1)代入b(n)的表达式
b(n)=1/(-n)[-(n+1)]=1/[n(n+1)]
1/nb(n)=n+1
∴{1/nb(n)}是首项为2, 公差为1的等比数列
Tn=2n+n(n-1)/2