Question

求数列通项公式

Answer 1

(1)
{ (1- Sn)/an } 是常数列
n=1
(1-a1)/a1 = C
(1-2/3)/(2/3) =C
C=1/2
ie
(1- Sn)/an =1/2
2-2Sn = an (1)
2-2S(n+1) = a(n+1) (2)
(2)-(1)
a(n+1)-an = -2a(n+1)
3a(n+1)= an
=>{ an} 是等比数列 , q=1/3
an = (1/3)^(n-1) .a1
an= 2.(1/3)^n
(2)
Sn
= a1+a2+...+an
=(2/3)( 1- 1/3^n) /(1- 1/3)
=1- 1/3^n
bn
= a(n+1)/[Sn.S(n+1)]
= 2.(1/3)^(n+1)/ [ (1- 1/3^n)(1- 1/3^(n+1)) ]
=2.3^n/ { (3^n- 1).[3^(n+1)- 1] }
=1/(3^n- 1) - 1/[3^(n+1)- 1]
Tn
=b1+b2+...+bn
=1/(3- 1) - 1/[3^(n+1)- 1]
=1/2 - 1/[3^(n+1)- 1]