1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+...+1/[(x+99)(x+100)]+1/(x+100)=199_数学解答

1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+...+1/[(x+99)(x+100)]+1/(x+100)=1999/2000

人气：384 ℃　时间：2020-09-15 05:41:38

解答

1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+...+1/[(x+99)(x+100)]+1/(x+100)
=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+99)-1/(x+100)+1/(x+100)
=1/(x+1)
=1999/2000
x=1/1999