数列 (24 10:6:29) 已知等差数列｛an｝的前n项和Sn,bn=1/Sn,且a3b3=1/2,S3+S5=211)求数列｛bn｝的通项公式2）求证：b1+b2+b3+...+bn＜2

数列 (24 10:6:29) 已知等差数列｛an｝的前n项和Sn,bn=1/Sn,且a3b3=1/2,S3+S5=211)求数列｛bn｝的通项公式2）求证：b1+b2+b3+...+bn＜2
数列 (24 10:6:29)
 
已知等差数列｛an｝的前n项和Sn,bn=1/Sn,且a3b3=1/2,S3+S5=21
1)求数列｛bn｝的通项公式
2）求证：b1+b2+b3+...+bn＜2

数列 (24 10:6:29) 已知等差数列｛an｝的前n项和Sn,bn=1/Sn,且a3b3=1/2,S3+S5=211)求数列｛bn｝的通项公式2）求证：b1+b2+b3+...+bn＜2
1)设an=a1+(n-1)d,则Sn=na1+n(n-1)d/2
由a3b3=1/2,bn=1/Sn,得S3=2a3=2a1+4d=3a1+3d,a1=d
所以Sn=n(n+1)d/2
由S3+S5=21,得6d+15d=21,d=1
所以Sn=n(n+1)/2
所以：bn=2/[n(n+1)]
2)b1+b2+b3+……+bn
=2{1/[1*2]+1/[2*3]+……+1/[n*(n+1)]}
=2[1-1/2+1/2-1/3+……+1/n-1/(n+1)]
=2[1-1/(n+1)]

a(n) = a + (n-1)d, n = 1,2,...
S(n) = na + n(n-1)d/2.
b(n) = 1/S(n) = 1/[na + n(n-1)d/2] = 2/[2na + n(n-1)d],
1/2 = a(3)b(3) = [a+2d]*2/[6a+6d] = (a+2d)/(3a+3d),
3a+3d = 2a + 4d, a = d...

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a(n) = a + (n-1)d, n = 1,2,...
S(n) = na + n(n-1)d/2.
b(n) = 1/S(n) = 1/[na + n(n-1)d/2] = 2/[2na + n(n-1)d],
1/2 = a(3)b(3) = [a+2d]*2/[6a+6d] = (a+2d)/(3a+3d),
3a+3d = 2a + 4d, a = d.
b(n) = 2/[2na + n(n-1)d] = 2/[n(n+1)d].
S(n) = na + n(n-1)d/2 = nd[2 + n-1]/2 = n(n+1)d/2.
21 = S(3)+S(5) = 3*4d/2 + 5*6d/2 = 6d + 15d = 21d, d = 1,
b(n) = 2/[n(n+1)d] = 2/[n(n+1)], n = 1,2,...
2.bn=2/[n(n+1)]
=2(1/n-1/(n+1)]
so b1+b2+b3...+bn
=2[1-1/2+1/2-1/3-----+1/n-1/(n+1)中间的都抵消
=2[1-1/(n+1)]=2-2/(n+1)<2得证

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a(n) = a + (n-1)d, n = 1,2,...
S(n) = na + n(n-1)d/2.
b(n) = 1/S(n) = 1/[na + n(n-1)d/2] = 2/[2na + n(n-1)d],
1/2 = a(3)b(3) = [a+2d]*2/[6a+6d] = (a+2d)/(3a+3d),
3a+3d = 2a + 4d, a = d...

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a(n) = a + (n-1)d, n = 1,2,...
S(n) = na + n(n-1)d/2.
b(n) = 1/S(n) = 1/[na + n(n-1)d/2] = 2/[2na + n(n-1)d],
1/2 = a(3)b(3) = [a+2d]*2/[6a+6d] = (a+2d)/(3a+3d),
3a+3d = 2a + 4d, a = d.
b(n) = 2/[2na + n(n-1)d] = 2/[n(n+1)d].
S(n) = na + n(n-1)d/2 = nd[2 + n-1]/2 = n(n+1)d/2.
21 = S(3)+S(5) = 3*4d/2 + 5*6d/2 = 6d + 15d = 21d, d = 1,
b(n) = 2/[n(n+1)d] = 2/[n(n+1)], n = 1,2,...

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b3= 1/S3
a3/S3=1/2
a3/(a1+a2+a3)=1/2 a1+a2=a3
a3=a2+d
a1=d
an=na1
S3+S5 =a1+2a1+3a1+a1+2a1+3a1+4a1+5a1=21
a1=1 an=n
Sn=(1+n)n/2
bn=2/[n(n+1)]
bn= 2[1/n-...

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b3= 1/S3
a3/S3=1/2
a3/(a1+a2+a3)=1/2 a1+a2=a3
a3=a2+d
a1=d
an=na1
S3+S5 =a1+2a1+3a1+a1+2a1+3a1+4a1+5a1=21
a1=1 an=n
Sn=(1+n)n/2
bn=2/[n(n+1)]
bn= 2[1/n- 1/(n+1)]
b1=2(1-1/2)
b2=2(1/2-1/3)
b3=2(1/3-1/4)
……
bn= 2[1/n- 1/(n+1)]
全部加起来
b1+b2+b3+...+bn= 2[1-1/(n+1)]=2-2/(n+2)<2
得证

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(1)
设首项是a1，公差是d
a3=a1+2d
S3=3a1+3d
∴b3=1/S3
=(a1+2d)/(3a1+3d)=1/2
∴2a1+4d=3a1+3d
∴a1=d
S3=3a1+3d=6a1
S5=5a1+(5*4/2)d=5a1+10d=15a1
∴S3+S5=21a1=21
∴a=d=1<...

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(1)
设首项是a1，公差是d
a3=a1+2d
S3=3a1+3d
∴b3=1/S3
=(a1+2d)/(3a1+3d)=1/2
∴2a1+4d=3a1+3d
∴a1=d
S3=3a1+3d=6a1
S5=5a1+(5*4/2)d=5a1+10d=15a1
∴S3+S5=21a1=21
∴a=d=1
∴Sn=na+[n(n-1)/2]d
=n+n(n-1)/2=(n^2+n)/2
∴bn=1/Sn=2/(n^2+n)
(2)
∵bn=2/(n^2+n)=2/[n(n+1)]
∴S(bn)=2/(1*2)+2/(2*3)+……+2/[n(n+1)]
=2*{(1/1-1/2)+(1/2-1/3)+……+[1/n-1/(n+1)]}
=2*[1-1/(n+1)]
=2n/(n+1)
∵n/(n+1)＜1
∴2n/(n+1)＜2
即：S(bn)＜2

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