数列1,1/(1+2),1/(1+2+3),..,1/(1+2+3+...n),...,则其前n项的和为49/25,则项数为

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数列1,1/(1+2),1/(1+2+3),..,1/(1+2+3+...n),...,则其前n项的和为49/25,则项数为
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数列1,1/(1+2),1/(1+2+3),..,1/(1+2+3+...n),...,则其前n项的和为49/25,则项数为
数列1,1/(1+2),1/(1+2+3),..,1/(1+2+3+...n),...,则其前n项的和为49/25,则项数为

数列1,1/(1+2),1/(1+2+3),..,1/(1+2+3+...n),...,则其前n项的和为49/25,则项数为
an=1/(1+2+...+n)=1/[n(n+1)/2]=2/n(n+1)=2(1/n-1/(n+1))
所以Sn=a1+a2+...+an
=2[(1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1))]
=2[1-1/(n+1)]
=2n/(n+1)
令Sn=2n/(n+1)=49/25
得n=49
  很高兴为你答疑,

令1/An=1+2+3+...+n=n(n+1)/2,则
An=2/n-2/(n+1)
Sn=2/1-2/2+2/2-2/3+2/3-2/4+...+ 2/n-2/(n+1)
=2-2/(n+1)=49/25
1-1/(n+1)=49/50
1/(n+1)=1/50
所以n=49