Question

求解答 需要过程

Answer 1

追答

Answer 2

(17)
(I)
a1=1
an =a1.q^(n-1) = q^(n-1)

for n>=3

an= [a(n-1)+ a(n-2)]/2

n=3
a3 = (a2+a1)/2
2q^2 =q+1

2q^2 -q-1=0
(2q-1)(q+1)=0
q=1/2 or -1

(II)
case 1: q=-1
an = (-1)^(n+1)
bn =nan
Sn = (-1)^(n+1) .n
= (n+1)/2 if n=odd
= -n/2 if n is even

case 2: q= 1/2
an = (1/2)^(n-1)
bn = n.(1/2)^(n-1)

let
S = 1.(1/2)^0+2.(1/2)^1+.....+n.(1/2)^(n-1) (1)
(1/2)S = 1.(1/2)^1+2.(1/2)^2+.....+n.(1/2)^n (2)

(1)-(2)
(1/2)S =[ 1+ 1/2+1/2^2+...+1/2^(n-1) ] - n.(1/2)^n
= 2( 1- (1/2)^n ) - n.(1/2)^n
S = 4 - 2(n+2).(1/2)^n

b1+b2+...+bn=S = 4 - 2(n+2).(1/2)^n