计算:1)1/(1x2)+1/(2x3)+1/(3x4)+……1/(n-1)n=
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计算:1)1/(1x2)+1/(2x3)+1/(3x4)+……1/(n-1)n=
2)1+1/91+2)+1/(1+2+3)+1/(1+2+3+4)+…+1/(1+2+3+…+n)=
2)1+1/91+2)+1/(1+2+3)+1/(1+2+3+4)+…+1/(1+2+3+…+n)=
答:
1)
1/(1x2)+1/(2x3)+1/(3x4)+……1/(n-1)n=
=1-1/2+1/2-1/3+1/3-1/4+.+1/(n-1)-1/n
=1-1/n (中间各项抵消了)
=(n-1)/n
2)
第n项的分母为自然数之和=(n+1)n/2
所以第n项为2/[(n+1)n]=2/n-2/(n+1)
所以:
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+…+1/(1+2+3+…+n)
=2×[1-1/2+1/2-1/3+1/3-1/4+.1/n-1/(n+1)]
=2×[1-1/(n+1)]
=2-2/(n+1)
=2n/(n+1)
1)
1/(1x2)+1/(2x3)+1/(3x4)+……1/(n-1)n=
=1-1/2+1/2-1/3+1/3-1/4+.+1/(n-1)-1/n
=1-1/n (中间各项抵消了)
=(n-1)/n
2)
第n项的分母为自然数之和=(n+1)n/2
所以第n项为2/[(n+1)n]=2/n-2/(n+1)
所以:
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+…+1/(1+2+3+…+n)
=2×[1-1/2+1/2-1/3+1/3-1/4+.1/n-1/(n+1)]
=2×[1-1/(n+1)]
=2-2/(n+1)
=2n/(n+1)
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