请阅读下列材料:∵1/1*3=1/2﹙1-1/3﹚,1/3*5=﹙1/3-1/5﹚···1/2009*2011=1/2﹙1/2009-1/2011﹚,∴1/1*3+1/3*5+1/5*7+···+1/2009*2011=1/2﹙1/1-1/3+1/3-1/5+1/5-1/7+···1/2009-1/2011﹚=1/2*﹙1-1/2011﹚=1005/2011在和式1/1*3+1/3*5+1/
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![请阅读下列材料:∵1/1*3=1/2﹙1-1/3﹚,1/3*5=﹙1/3-1/5﹚···1/2009*2011=1/2﹙1/2009-1/2011﹚,∴1/1*3+1/3*5+1/5*7+···+1/2009*2011=1/2﹙1/1-1/3+1/3-1/5+1/5-1/7+···1/2009-1/2011﹚=1/2*﹙1-1/2011﹚=1005/2011在和式1/1*3+1/3*5+1/](/uploads/image/z/4069009-1-9.jpg?t=%E8%AF%B7%E9%98%85%E8%AF%BB%E4%B8%8B%E5%88%97%E6%9D%90%E6%96%99%3A%E2%88%B51%2F1%2A3%3D1%2F2%EF%B9%991-1%2F3%EF%B9%9A%2C1%2F3%2A5%3D%EF%B9%991%2F3-1%2F5%EF%B9%9A%C2%B7%C2%B7%C2%B71%2F2009%2A2011%3D1%2F2%EF%B9%991%2F2009-1%2F2011%EF%B9%9A%2C%E2%88%B41%2F1%2A3%2B1%2F3%2A5%2B1%2F5%2A7%2B%C2%B7%C2%B7%C2%B7%2B1%2F2009%2A2011%3D1%2F2%EF%B9%991%2F1-1%2F3%2B1%2F3-1%2F5%2B1%2F5-1%2F7%2B%C2%B7%C2%B7%C2%B71%2F2009-1%2F2011%EF%B9%9A%3D1%2F2%2A%EF%B9%991-1%2F2011%EF%B9%9A%3D1005%2F2011%E5%9C%A8%E5%92%8C%E5%BC%8F1%2F1%2A3%2B1%2F3%2A5%2B1%2F)
请阅读下列材料:∵1/1*3=1/2﹙1-1/3﹚,1/3*5=﹙1/3-1/5﹚···1/2009*2011=1/2﹙1/2009-1/2011﹚,∴1/1*3+1/3*5+1/5*7+···+1/2009*2011=1/2﹙1/1-1/3+1/3-1/5+1/5-1/7+···1/2009-1/2011﹚=1/2*﹙1-1/2011﹚=1005/2011在和式1/1*3+1/3*5+1/
请阅读下列材料:
∵1/1*3=1/2﹙1-1/3﹚,1/3*5=﹙1/3-1/5﹚···1/2009*2011=1/2﹙1/2009-1/2011﹚,
∴1/1*3+1/3*5+1/5*7+···+1/2009*2011=1/2﹙1/1-1/3+1/3-1/5+1/5-1/7+···1/2009-1/2011﹚=1/2*﹙1-1/2011﹚=1005/2011
在和式1/1*3+1/3*5+1/5*7+···中第5项为___,第n项为___.上述和的方法是将和式中的各项转化为两个数之差,使得首末两项外的中间各项可以___,从而打到目的.
利用上述结论计算1/x﹙x+2﹚+1/﹙x+2﹚﹙x+4﹚+1/﹙c+a﹚﹙x+6﹚+···+1/﹙x+2010﹚﹙x+2012﹚
请阅读下列材料:∵1/1*3=1/2﹙1-1/3﹚,1/3*5=﹙1/3-1/5﹚···1/2009*2011=1/2﹙1/2009-1/2011﹚,∴1/1*3+1/3*5+1/5*7+···+1/2009*2011=1/2﹙1/1-1/3+1/3-1/5+1/5-1/7+···1/2009-1/2011﹚=1/2*﹙1-1/2011﹚=1005/2011在和式1/1*3+1/3*5+1/
(1) 在和式1/1*3+1/3*5+1/5*7+···中第5项为_1/9*11_,第n项为_1/(2n-1)*(2n+1)__.上述和的方法是将和式中的各项转化为两个数之差,使得首末两项外的中间各项可以_抵消__,从而打到目的.
(2) 1/﹙c+a﹚﹙x+6﹚是不是打错了 应该是1/﹙x+4﹚﹙x+6)吧 结果是1016/x(x+2012)