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

题目

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

答案

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)