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
1个回答
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
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