Question

数列{an}中,a3=1,a1+a2+...+an=a(n+1)(n=1,2,3...)

1)求a1，a2
2）求Sn和an
3）设bn=log2Sn，有{cn}使得cn*b(n+3)*b(n+4)=1+n(n+1)(n+2)Sn，求cn的前n项和Tn

Answer 1

a(1)+a(2)+...+a(n) = a(n+1),
a(n+2) = a(1)+a(2)+...+a(n)+a(n+1) = a(n+1) + a(n+1) = 2a(n+1),
a(1+2) = a(3) = 1 = 2a(1+1) = 2a(2), a(2) = 1/2.
a(3) = 1 = a(1) + a(2) = a(1) + 1/2, a(1) = 1/2.

{a(n+1)}是首项为a(2)=1/2, 公比为2的等比数列。
a(n+1) = (1/2)*2^(n-1) = 2^(n-2),

a(1)=1/2,
n>=2时，a(n)= 2^(n-3).

s(n) = a(1)+a(2)+...+a(n) = a(n+1) = 2^(n-2).

b(n)=log_{2}[s(n)] = log_{2}[2^(n-2)] = n-2,

1 + n(n+1)(n+2)s(n) = 1 + n(n+1)(n+2)2^(n-2) = c(n)*b(n+3)b(n+4) = (n+1)(n+2)c(n),
c(n) = 1/[(n+1)(n+2)] + n2^(n-2)
= 1/(n+1) - 1/(n+2) + d(n),

d(n) = n2^(n-2),
D(n) = d(1)+d(2)+d(3)+...+d(n-1)+d(n)
=2^(1-2) + 2*2^(2-2) + 3*2^(3-2) + ... + (n-1)2^(n-3) + n2^(n-2),
2D(n) = 2^(2-2) + 2*2^(3-2) + ... + (n-1)2^(n-2) + n2^(n-1),

D(n) = 2D(n)-D(n) = -2^(1-2) - 2^(2-2) - 2^(3-2) - ... - 2^(n-2) + n2^(n-1)
= n2^(n-1) - 2^(-1)[1 + 2 + ... + 2^(n-1)]
= n2^(n-1) - (1/2)[2^n - 1]/(2-1)
= n2^(n-1) - [2^n - 1]/2
= (n-1)2^(n-1) + 1/2,

t(n) = c(1)+c(2)+...+c(n-1)+c(n)
= [1/2-1/3 + 1/3-1/4 + ... + 1/n-1/(n+1) + 1/(n+1)-1/(n+2)] + D(n)
= 1/2 - 1/(n+2) + (n-1)2^(n-1) + 1/2
= 1 - 1/(n+2) + (n-1)2^(n-1)
= (n+1)/(n+2) + (n-1)2^(n-1)

Answer 2

（1）、有题意得知a1=a2 a1+a2=a3=1
所以a1=a2=1/2

(2)、由题意得：an=a1+a2+...+a(n-1)(n>=2,n属于正整数)
a(n+1)/an=[a1+a2+...+a(n-1)+an]/an=2an/an=2, (n>=2,n属于正整数)
所以数列{an}为公比为2的等比数列，(n>=2,n属于正整数)
当n=1是S1=a1=1/2
an=1/2*2^(n-2)=2^(n-3) ，(n>=2,n属于正整数)
Sn=a(n+1)=2an=2^(n-2) ，(n>=2,n属于正整数)
当n=1时S1=2^(1-2)=1/2
所以Sn=Sn=a(n+1)=2an=2^(n-2) ，(n>=1,n属于正整数)

(3)、因为bn=log2Sn=log2 2^(n-2)=n-2
cn*b(n+3)*b(n+4)=1+n(n+1)(n+2)Sn
cn*(n+1)*(n+2)=1+n(n+1)(n+2)2^(n-2) 两边同除(n+1)(n+2)
cn=1/[(n+1)(n+2)]+n*2^(n-2) 接着就用列项相消法和错位相减法来求cn的前n项和Tn

Tn=1/[(1+1)(1+2)]+1/[(2+1)(2+2)]+.....+1/[(n)(n+1)] +1/[(n+1)(n+2)]
+1*2^(1-2) +2*2^(2-2) +......+(n-1)*2^(n-3) +n*2^(n-2)

设Tn=An+Bn
An=1/[(1+1)(1+2)]+1/[(2+1)(2+2)]+.....+1/[(n)(n+1)] +1/[(n+1)(n+2)]
Bn=1*2^(1-2) +2*2^(2-2) +......+(n-1)*2^(n-3) +n*2^(n-2)
因为1/[(n+1)(n+2)]=1/(n+1)-1/(n+2)
所以An=1/2-1/3+1/3-1/4+....+1/n-1/(1+n)+1/(1+n)-1/(2+n)=1/2-1/(2+n)

1) Bn=1*2^(1-2)+2*2^(2-2) +......+(n-1)*2^(n-3) +n*2^(n-2)
2) 2 Bn= 1*2^(2-2)+2*2^(3-2) +...... + (n-1)*2^(n-2) +n*2^(n-1)
1)-2)得-Bn=2^(-1)+2+2^2+....+2^(n-2)-n*2^(n-1)
=[1/2(1-2^n)]/(1-2)-n*2^(n-1)
=2^(n-1)-1/2-n*2^(n-1)
=(1-n)*2^(n-1)-1/2
Bn=(n-1)*2^(n-1)+1/2
所以Tn=An+Bn
=1/2-1/(2+n)+(n-1)*2^(n-1)+1/2
=(n-1)*2^(n-1)+(n+1)/(n+2)