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计算x(x+1)分之1+(x+1)(x+2)分之1+(x+2)(x+3)分之1+……+(x+2011)(x+2012)分之1
人气:292 ℃ 时间:2020-04-27 13:17:01
解答
裂项相消:
1/[x(x+1)]+1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+...+1/[(x+2011)(x+2012)]
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+2011)-1/(x+2012) 中间的都消掉了
=1/x-1/(x+2012)
=(x+2012-x)/[x(x+2012)]
=2012/[x(x+2012)]
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