已知an,bn满足a1=2,b1=1且an=3/4a(n-1)+1/4b(n-1)+1,bn=1/4a(n-1)+3/4
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已知an,bn满足a1=2,b1=1且an=3/4a(n-1)+1/4b(n-1)+1,bn=1/4a(n-1)+3/4b(n-1)+1(a>=2) 令cn=an+bn 设数列Cn的前n和为sn,求证1/S1+1/S2+1/S3+.+1/Sn
n要大于等于2
n要大于等于2
Cn=an+bn
=[3/4a(n-1)+1/4b(n-1)+1] + [1/4a(n-1)+3/4b(n-1)+1]
= a(n-1)+b(n-1) +2
=C(n-1) +2
C1=a1+b1=3
所以Cn是等差数列,Cn=2n+1
Sn= 3+5+7+……+(2n+1)
= n+n(n+1) = n(n+2)
1/Sn= 1/[n(n+2)]
= [1/n - 1/(n+2)]/2
1/S1+1/S2+1/S3+.+1/Sn = (1- 1/3)/2 + (1/2 - 1/4)/2 + (1/3 - 1/5)/2+.+ [1/n - 1/(n+2)]/2
= [ 3/2 - 1/(n+1) - 1/(n+2)] / 2
< 3/4
=[3/4a(n-1)+1/4b(n-1)+1] + [1/4a(n-1)+3/4b(n-1)+1]
= a(n-1)+b(n-1) +2
=C(n-1) +2
C1=a1+b1=3
所以Cn是等差数列,Cn=2n+1
Sn= 3+5+7+……+(2n+1)
= n+n(n+1) = n(n+2)
1/Sn= 1/[n(n+2)]
= [1/n - 1/(n+2)]/2
1/S1+1/S2+1/S3+.+1/Sn = (1- 1/3)/2 + (1/2 - 1/4)/2 + (1/3 - 1/5)/2+.+ [1/n - 1/(n+2)]/2
= [ 3/2 - 1/(n+1) - 1/(n+2)] / 2
< 3/4
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