Question

已知Sn是数列{an}的前n项和,Sn满足关系式2Sn=S(n-1)-【(1/2)^(n-1)】+2 (n>=2,n为正

Answer 1

(1)

2Sn=S(n-1)-【(1/2)^(n-1)】+2

2^n. Sn - 2^(n-1) S(n-1) =-1+ 2^(n-1)

2^n. Sn - 2^1. S1 =(-1+ 2^1)+(-1+2^2)+...+[-1+ 2^(n-1)]

= -(n-1) + 2[ 2^(n-1) -1 ]

= -n-1 + 2^n

2^n. Sn = -n + 2^n

Sn = 1-n.(1/2)^n

an = Sn - S(n-1)

= [ -n/2 + (n-1) ](1/2)^(n-1)

=[ (n-2)/2] . (1/2)^(n-1) ; n>=2

= 1/2 ; n=1

(2)

Sn = 1-n.(1/2)^n

consider

f(x) = 1- x(1/2)^x

f'(x) = -(1/2)^x. [ 1+ x ln(1/2) ] =0

x = -1/ln(1/2) = 1.442

S1 = 1/2

S2 = 1-2.(1/2)^2 = 1/2

Sn的取值范围 = [ 1/2 , 1)