已知数列1/1,1/(1+2),1/(1+2+3),1/(1+2+3+4)…的前n项和为20/11,则项数n为
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![已知数列1/1,1/(1+2),1/(1+2+3),1/(1+2+3+4)…的前n项和为20/11,则项数n为](/uploads/image/z/9633798-54-8.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%971%2F1%2C1%2F%281%2B2%29%2C1%2F%281%2B2%2B3%29%2C1%2F%281%2B2%2B3%2B4%29%E2%80%A6%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BA20%2F11%2C%E5%88%99%E9%A1%B9%E6%95%B0n%E4%B8%BA)
已知数列1/1,1/(1+2),1/(1+2+3),1/(1+2+3+4)…的前n项和为20/11,则项数n为
已知数列1/1,1/(1+2),1/(1+2+3),1/(1+2+3+4)…的前n项和为20/11,则项数n为
已知数列1/1,1/(1+2),1/(1+2+3),1/(1+2+3+4)…的前n项和为20/11,则项数n为
易发现数列的通项公式为:an=1/[n(n+1)/2]=2/n(n+1)=2[1/n-1/(n+1)]
前n项和Sn=2[1-1/2+1/2-1/3+...+1/(n-1)-1/n]
=2(1-1/n)
由题20/11=2(1-1/n) 解得n=11
n=10.通项公式为2/(n(n+1))=2(1/n-1/(n+1)),前n项和2(1-1/(n+1))