(i)
a(n+1)=2/(an+1)
a(n+1) -1 =2/(an+1) -1
= -(an-1)/(an+1)
1/[a(n+1) -1] = -(an+1)/(an-1)
= -1 - 2/(an-1)
1/[a(n+1) -1]+1/3 = -2[ 1/(an-1) +1/3]
=> {1/(an-1) +1/3} 是等比数列,q=-2
1/(an-1) +1/3 = (-2)^(n-1).(1/(a1-1) +1/3)
=-(2/3)(-2)^n
1/(an-1) = -1/3 -(2/3)(-2)^n
an = 1- 1/[1/3 +(2/3)(-2)^n]
an +2 = 3 - 1/[1/3 +(2/3)(-2)^n]
= 3 -3/[1 +2.(-2)^n] (1)
= 6.(-2)^n/[1 +2.(-2)^n]
an -1 = - 1/[1/3 +(2/3)(-2)^n]
= -3/[1 +2.(-2)^n] (2)
(1)/(2)
=>(an+2)/(an-1) = (-2)^(n+1)
bn = |(an+2)/(an-1)| = 2^(n+1)
(ii)
Sn = a1+a2+...+an
Sn = (9/8)an-(1/2)*3^n+0.5
n=1,a1=8
an = Sn -S(n-1)
=(9/8)an -(9/8)a(n-1) - 3^(n-1)
an = 9a(n-1) +8.3^(n-1)
an + (4/3).3^n = 9[ a(n-1) + (4/3).3^(n-1) ]
{an + (4/3).3^n}是等比数列,q=9
an + (4/3).3^n = 9^(n-1) .(a1 + (4/3).3^1)
= 12.9^(n-1)
an = -(4/3).3^n +12.9^(n-1)
Sn = -2( 3^n-1) + (3/2)(9^n -1)
= (3/2)9^n - 2.3^n + 1/2
= (1/2) [ 3(3^n)^2- 4(3^n) + 1]
= (1/2) (3^(n+1) -1)(3^n-1)
bn = 3^n/Sn
= 2.3^n/[(3^(n+1) -1)(3^n-1)]
= 1/(3^n-1) - 1/(3^(n+1) -1)
Tn = b1 +b2+...+bn
= 1/2 -1/(3^(n+1) -1)