关于分式计算的题2/(x+1)(x+3)+2/(x+3)(x+5)+…+2/(x+99)(x+101)=?_数学解答

关于分式计算的题
2/(x+1)(x+3)+2/(x+3)(x+5)+…+2/(x+99)(x+101)=?

人气：462 ℃　时间：2020-02-03 07:18:14

解答

=2[1/(x+1)(x+3)+1/(x+3)(x+5)+…+1/(x+99)(x+101)]
=2{[2/(X+1)(X+3)+2/(X+3)(X+5)+2/(X+5)(X+7)+...+2/(X+99)(X+101)]/2}
=1/(X+1)-1/(X+3)+1/(X+3)-1/(X+5)+1/(X+5)-1/(X+7)+...+1/(X+99)-1/(X+101)
=1/(X+1)-1/(X+101)
=(X+101-X-1)/(X+1)(X+101)
=100/(X+1)(X+101)