Question

求（2）第二小问过程？

我算的an＝1/2^n bn＝n-1 Tn＝1/2^n+n-1【可能不对】

Answer 1

(1)
an =a1.q^(n-1) = (1/2).q^(n-1) ; q>0
Sn=a1+a2+...+an
bn=b1+(n-1)d
1/a3= 1/a2 +4
1/[(1/2)q^2] = 1/[(1/2)q] + 4
2/q^2 = 2/q +4
2q^2 +q -1 =0
(2q-1)(q+1)=0
q=1/2
an = (1/2)^n
a3 =1/(b4+b6)
(1/2)^3 = 1/( 2b1 +8d)
2b1+8d =8
b1+4d =4 (1)
a4= 1/(b5+2b7)
1/16 = 1/(3b1+16d)
3b1+16d =16 (2)
4(1)-(2)
b1 =0
from (1)
d=1
bn = n-1
(2)
Sn = a1+a2+...+an = 1 - (1/2)^n
Tn
= S1+S2+...+Sn
= n - ( 1- 1/2^n )
=n-1 + (1/2)^n
To prove : ∑(i:1->n) [T(i+1) -b(i+1)].b(i+3) ]/[b(i+1).b(i+2) ] <1/2
∑(i:1->n) [T(i+1) -b(i+1)].b(i+3) ]/[b(i+1).b(i+2) ]
=∑(i:1->n) [ i +(1/2)^(i+1) -i ]. (i+2) ]/[i(i+1) ]
=∑(i:1->n) (i+2) (1/2)^(i+1)/[i(i+1) ]
=∑(i:1->n) [ 2/i - 1/(i+1) ] (1/2)^(i+1)
=∑(i:1->n) [ (1/2)^i/i - (1/2)^(i+1) /(i+1)]
= (1/2)/1 - (1/2)^(n+1)/(n+1)
= 1/2 - (1/2)^(n+1)/(n+1)
< 1/2