Question

已知数列｛an｝是公比大于1的等比数列,对任意的n∈N*有,an+1=a1+a2+...+an-1+5/2an+1/2

Answer 1

an=a1.q^(n-1)

a(n+1)=a1+a2+...+a(n-1)+(5/2)an+1/2

n=1,

a2=(5/2)a1+1/2

a1q =(5/2)a1+1/2 (1)

n=2,

a3=a1+(5/2)a2+1/2

a1q^2=a1+(5/2)a1q+1/2 (2)

(2)-(1)

q(q-1) =-(3/2) +(5/2)q

2q^2-7q+3=0

(2q-1)(q-3)=0

q=3 (q>1)

from (1)

a1q =(5/2)a1+1/2

3a1=(5/2)a1+1/2

a1=1

ie

an= 3^(n-1)

(2)

bn=b1+(n-1)d

bn=(1/n)[ log(a1)+log(a2)+...+log(an)+log(t) ]

=(1/n)[ n(n-1)/2 +log(t) ]

n=1,

b1 =log(t) (3)

n=2,

b2=(1/2)[ 1 +log(t) ]

b1+d =(1/2)[ 1 +log(t) ] (4)

(4)-(3)

d= (1/2)[ 1 -log(t) ]

n=3,

b3=(1/3)[ 3 +log(t) ]

b1+2d=(1/3)[ 3 +log(t) ] (5)

(5)-(4)

d = 1/2 -(1/6)log(t)

(1/2)[ 1 -log(t) ] =1/2 -(1/6)log(t)

t=1

d=1/2

b1=0

bn= (n-1)/2