先证明了左右两边首尾两项的等量关系：
2/(x^2-1)+20/(x^2-100)=(22x^-220)/[(x^2-1)(x^2-100)]
=(22x^-220)/[(x+1)(x-1)(x+10)(x-10)]
=22(x^-10)/[(x+1)(x-1)(x+10)(x-10)]
11/(x-1)(x+10)+11/(x-10)(x+1)=11(2x^2+20)/[(x+1)(x-1)(x+10)(x-10)]
=22(x^2+10)/[(x+1)(x-1)(x+10)(x-10)]
即2/(x^2-1)+20/(x^2-100)=11/(x-1)(x+10)+11/(x-10)(x+1)
同理4/(x^2-4)+18/(x^2-81)=11/(x-2)(x+9)+11/(x-9)(x+2)
……
所以：2/(x2-1)+4/(x2-4)+6/(x2-9)+…+20/(x2-100)=11/(x-1)(x+10)+…+11/(x-10)(x+1)