数列1,1/(1+2),1/(1+2+3),...,1/(1+2+3+...+n),...的前n项和为49/25,则项数
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数列1,1/(1+2),1/(1+2+3),...,1/(1+2+3+...+n),...的前n项和为49/25,则项数{an}为
![数列1,1/(1+2),1/(1+2+3),...,1/(1+2+3+...+n),...的前n项和为49/25,则项数](/uploads/image/z/18961330-58-0.jpg?t=%E6%95%B0%E5%88%971%2C1%2F%281%2B2%29%2C1%2F%281%2B2%2B3%29%2C...%2C1%2F%281%2B2%2B3%2B...%2Bn%29%2C...%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BA49%2F25%2C%E5%88%99%E9%A1%B9%E6%95%B0)
an=1/[n(n+1)/2]=2/n(n+1)=2/n-2/(n+1)
a1+a2+...+an=2/1-2/2+2/2-2/3+...+2/n-2/(n+1)=2/1-2/(n+1)=49/25
解得n=49,an=1/[n(n+1)/2]=1/1225
a1+a2+...+an=2/1-2/2+2/2-2/3+...+2/n-2/(n+1)=2/1-2/(n+1)=49/25
解得n=49,an=1/[n(n+1)/2]=1/1225
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