若n为正整数,求1/n(n+1)+1/(n+1)(n+2)＋1/(n+2)(n+3)+1/(n+3)(n+4)+.+1/(n+99)(n+100)的值观察算式1/1X2=1-1/2 1/1x2+1/2x3=1-1/2+1/2-1/3=2/3.

拆项
1/n(n+1)+1/(n+1)(n+2)＋1/(n+2)(n+3)+1/(n+3)(n+4)+....+1/(n+99)(n+100)
=[1/n-1/(n+1)]+[1/(n+1)-1/(n+2)]+....+[1/(n+99)-1/(n+100)]
=1/n-1/(n+1)+1/(n+1)-1/(n+2)+1/(n+2)+...1/(n+99)-1/(n+100)
前后抵消得
=1/n-1/(n+100)
=100/[n(n+100)]
=100/(n^2+100n)

1/n(n+1)+1/(n+1)(n+2)＋1/(n+2)(n+3)+1/(n+3)(n+4)+....+1/(n+99)(n+100)
=1/n-1/(n+1)+1/(n+1)-1/(n+2)+...+1/(n+99)-1/(n+100)
=1/n-1/(n+100)
=100/n(n+100)

1/n(n+1)+1/(n+1)(n+2)＋1/(n+2)(n+3)+1/(n+3)(n+4)+.+1/(n+99)(n+100)
=1/n-1/(n+1)+1/(n+1)-1/(n+2)+...+1/(n+99)-1/(n+100)
=1/n-1/(n+100)
=100/n(n+100)