
2个回答
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an=1+2+3+……+n
=n(1+n)/2
1/an=2/[n(n+1)]
=2[1/n - 1/(n+1)]
数列{1/an}的前n项和Sn=a1+a2+a3+……+an
=2[1/1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)
Sn=2n/(n+1) (n为正整数)
=n(1+n)/2
1/an=2/[n(n+1)]
=2[1/n - 1/(n+1)]
数列{1/an}的前n项和Sn=a1+a2+a3+……+an
=2[1/1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)
Sn=2n/(n+1) (n为正整数)
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