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1+2+3+...+n=n(n+1)/2
1/(1+2+3+...+n)=1/[n(n+1)/2]=2/n(n+1)=2[1/n-1/(n+1)]
于是
原式
=2[1/1-1/(1+1)]+2[1/2-1/(2+1)]+2[1/3-1/(3+1)]+……+2[1/n-1/(n+1)]
=2[1/1-1/2]+2[1/2-1/3]+2[1/3-1/4]+……+2[1/n-1/(n+1)]
=2[1/1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1)]
=2[1/1-1/(n+1)
=2-2/(n+1)]
=2n/(n+1)
1/(1+2+3+...+n)=1/[n(n+1)/2]=2/n(n+1)=2[1/n-1/(n+1)]
于是
原式
=2[1/1-1/(1+1)]+2[1/2-1/(2+1)]+2[1/3-1/(3+1)]+……+2[1/n-1/(n+1)]
=2[1/1-1/2]+2[1/2-1/3]+2[1/3-1/4]+……+2[1/n-1/(n+1)]
=2[1/1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1)]
=2[1/1-1/(n+1)
=2-2/(n+1)]
=2n/(n+1)
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