数列求和 1+1/(1+2)+1/(1+2+3)+.+1/(1+2+3+.+n)
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数列求和 1+1/(1+2)+1/(1+2+3)+.+1/(1+2+3+.+n)
数列求和 1+1/(1+2)+1/(1+2+3)+.+1/(1+2+3+.+n)
数列求和 1+1/(1+2)+1/(1+2+3)+.+1/(1+2+3+.+n)
1/(1+2+3+……+n)=1/[n(n+1)/2]=2/[n(n+1)]
所以原式=1+2/2*3+2/3*4+……+2/[n(n+1)]
=1+2*[(1/2-1/3)+(1/3-1/4)+……+(1/n-1/(n+1)]
=1+2*[1/2-1/(n+1)]
=2-2/(n+1)
1+2(1/2-1/3)+2(1/3-1/4)+...+2(1/(n-1)-1/n)
=1+2(1/2-1/n)
设每一项为An
则an=1/(1+2+...n)=2/[n*(n+1)]
所以原式=2/(1*2)+2/(2*3)+2/(3*4)+......+2/(n*(n+1))
=2[1-1/2+1/2-1/3+1/3-1/4+........+n/1-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)
1+2+3+...+n=(n+1)*n/2,
1+1/(1+2)+1/(1+2+3)+......+1/(1+2+3+......+n)
=1+1/[(1+2)*2/2]+1/[(1+3)*3/2]+...+1/[(1+n)*n/2]
=1+2/(2*3)+2/(3*4)+...+2/[n(n+1)]
=1+2[1/(2*3)+1/(3*4)+...+1/n(n+1...
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1+2+3+...+n=(n+1)*n/2,
1+1/(1+2)+1/(1+2+3)+......+1/(1+2+3+......+n)
=1+1/[(1+2)*2/2]+1/[(1+3)*3/2]+...+1/[(1+n)*n/2]
=1+2/(2*3)+2/(3*4)+...+2/[n(n+1)]
=1+2[1/(2*3)+1/(3*4)+...+1/n(n+1)]
=1+2[(1/2)-(1/3)+(1/3)-(1/4)+...+(1/n)-1/(n+1)]
=1+2[(1/2)-1/(n+1)]
=1+2(n-1)/2(n+1)
=1+(n-1)/(n+1)
=(n+1+n-1)/(n+1)
=2n/(n+1).
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