设无穷数列｛an｝的前n项和为Sn.,数列｛Sn｝的前n项和为Cn,且Sn+Cn=n,求证数列｛1－Sn｝为等比数列,并求出an的通项公式

Question

设无穷数列｛an｝的前n项和为Sn.,数列｛Sn｝的前n项和为Cn,且Sn+Cn=n,
求证数列｛1－Sn｝为等比数列,并求出an的通项公式

Answer 1

1.求a1
S1 + C1 = 1
C1 = S1 = a1 = 0.5
2.求数列规律
Sn + Cn = Sn + (Sn + Sn-1 + Sn-2 + .+S2 +S1)
Sn-1 + Cn-1 = Sn-1 +(Sn-1 + Sn-2 + .+S2 +S1)
Sn + Cn = n
Sn-1 + Cn-1 = n-1
Sn + Cn - ( Sn-1 + Cn-1 ) =
Sn + (Sn - Sn-1) =
Sn + an = n - (n-1) = 1
则Sn+an = 1
且 1-Sn = an
Sn = 1- an
3.求 an
由a1 = 0.5 ,Sn = 1- an
Sn = a1 + a2 + .+an-1 + an
Sn-1 = a1 + a2 + .+an-1
则 Sn - Sn-1 = an
且 Sn = 1- an
Sn - Sn-1 = 1- an - ( 1 - an-1 ) = an-1 - an
很明显 Sn - Sn-1 = an = an-1 - an
an + an = an-1
则 an = an-1*0.5 ,且 a1 = 0.5 {an}这还不是等比数列?
an = 0.5* ( 0.5^(n-1) ) = 0.5^n
4.证明 {1-Sn} 等比数列
1-Sn = an = 0.5^n
{an} 为等比数列 ,
则 {1-Sn} 为等比数列
an = 0.5^n