已知等差数列{an}的前n项和为Sn,bn=1/Sn,且a3·b3=1/2,S3+S5=21

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已知等差数列{an}的前n项和为Sn,bn=1/Sn,且a3·b3=1/2,S3+S5=21
①求数列{bn}的通项公式
②求证b1+b2+...+bn

a(n) = a1 + (n-1)d,
S(n) = na1 + n(n-1)d/2.
b(n) = 1/S(n)
= 1/[na1 + n(n-1)d/2]
= 2/[2na1 + n(n-1)d],
1/2 = a3b3
= [a1+2d]*2/[6a1+6d]
= (a1+2d)/(3a1+3d),
3a1+3d = 2a1 + 4d,
a1=d.
b(n) = 2/[2na1 + n(n-1)d] = 2/[n(n+1)d].
S(n) = na1 + n(n-1)d/2
= nd[2 + n-1]/2
= n(n+1)d/2.
21 = S3+S5
= 3*4d/2 + 5*6d/2
= 6d + 15d
= 21d,
d = 1,
bn = 2/[n(n+1)d]
= 2/[n(n+1)],
b1=2/(1*2)
b2=2/(2*3)
b3=2/(3*4)
.
bn= 2/[n(n+1)],
b1+b2+...+bn
=2/(1*2)+2/(2*3)+2/(3*4)+.+2/[n(n+1)]
=2{1/(1*2)+1/(2*3)+1/(3*4)+.+1/[n(n+1)]}
=2{1-1/2+1/2-1/3+1/3-1/4+.+1/n-1/(n+1)}
=2[1-1/(n+1)]
=2-2/(n+1)

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