求证:2/x2-1+4/x2-4+6/x2-9+…+20/x2-100=11/(x-1)(x+10)+…+11/(x-1
求证:2/x2-1+4/x2-4+6/x2-9+…+20/x2-100=11/(x-1)(x+10)+…+11/(x-10)(x+1)
赞 卡卡婆娘 幼苗
共回答了27个问题采纳率：81.5% 举报
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)
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)
1年前
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