Question

一道初二数学计算题 3/x(x+2)+3/(x+2)(x+4)+3/(x+4)(x+6)+.+3/(x+20)(x+22

Answer 1

3/[x(x+2)]=3/2*[1/x-1/(x+2)]

3/[(x+2)(x+4)]=3/2*[1/(x+2)-(1/(x+4)]

.

3/[(x+20)(x+22)]=3/2*[1/(x+20)-1/(x+22)]

把上面的式子相加：

3/x(x+2)+3/(x+2)(x+4)+3/(x+4)(x+6)+.+3/(x+20)(x+22)

=3/2[1/x-1/(x+22)]

=33/[x(x+22)]