计算:1)1/(1x2)+1/(2x3)+1/(3x4)+……1/(n-1)n=
2个回答

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)