观察下面一组数-1/2,3/2,-5/2,7/2,-9/2,11/2……1.写出第2006个数；2.求这2006个数的和....
观察下面一组数-1/2,3/2,-5/2,7/2,-9/2,11/2……1.写出第2006个数；2.求这2006个数的和.
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展开全部1、第2006个数为4011/2.分析：从这些数可以看出每个数的分子都比前一个多2,分母不变!当数的个数是单数时,该数是负数!可一得出当n为奇数时,a[n]=-(2n-1).当n为偶数时,a[n]=(2n-1).所以当n=2006时,a[2006]=2×2006-1=4011.2、这2006个数的和是1003.分析：从数中可一看处a[1]+a[2]=-1/2+3/2=1=a[3]+a[4]=-5/2+7/2=1,若把a[n]+a[n+1]看为一组数,则2006个数共有这样的数共1003组,所以S[2006]=(a[1]+a[2])×1003=1003.([]中的是下角标!)有问题可以找我,很高兴能帮到你!',getTip:function(t,e){return t.renderTip(e.getAttribute(t.triangularSign),e.getAttribute("jubao"))},getILeft:function(t,e){return t.left+e.offsetWidth/2-e.tip.offsetWidth/2},getSHtml:function(t,e,n){return t.tpl.replace(/\{\{#href\}\}/g,e).replace(/\{\{#jubao\}\}/g,n)}},baobiao:{triangularSign:"data-baobiao",tpl:'{{#baobiao_text}}',getTip:function(t,e){return t.renderTip(e.getAttribute(t.triangularSign))},getILeft:function(t,e){return t.left-21},getSHtml:function(t,e,n){return t.tpl.replace(/\{\{#baobiao_text\}\}/g,e)}}};function l(t){return this.type=t.type
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观察按下列规则排成的一列数：1/1，1/2，2/1，1/3，2/2，3/1，1/4，2/3，3/2，4/1，1/5，2/4，3/3，4/2，，5/1，1/6，……（*）(...
观察按下列规则排成的一列数： 1/1，1/2，2/1，1/3，2/2，3/1，1/4，2/3，3/2，4/1，1/5，2/4，3/3，4/2，，5/1，1/6，……（*） (1)在(*)中,从左起第M个数记为F(M),当F(M)=2/2013时,求M的值和这M个数的积. (2)在(*)中,未经约分且分母为2的数记为C,它后面的一个数记为D,是否存在这样的两个数C和D,使CD=2001000,如果存在,求出C和D;如果不存在,说明理由.
速度 需要过程 谢谢
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m=2013×2014/2+2=2027093因为将上面的数分为n组，规律为分子是[1，2，3……n]，分母是[n，n-1，n-2……，3，2，1]，而且在每个分组内，相对应的分子与分母只和必为n+1，所以2013+2=n+1，n=2014，所以m=1+2+3+4+……+2013+2=2027093至于这些数的积很简单，每个分组内的所以数相乘为1，最后只剩1/2014×2/2013=1/2027091至于CD怎么求，同又设一个分组数n，则2001000=n×（n-1）/2，可求得n=2001，则C=2000/2,D=2001

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展开全部-n+1/n=-1+1/n所以随着n的增大，1/n变小，-1+1/n变小即-1/2与-2/3,-2/3与-3/4,-3/4与-4/5,-4/5与-5/6逐渐变小',getTip:function(t,e){return t.renderTip(e.getAttribute(t.triangularSign),e.getAttribute("jubao"))},getILeft:function(t,e){return t.left+e.offsetWidth/2-e.tip.offsetWidth/2},getSHtml:function(t,e,n){return t.tpl.replace(/\{\{#href\}\}/g,e).replace(/\{\{#jubao\}\}/g,n)}},baobiao:{triangularSign:"data-baobiao",tpl:'{{#baobiao_text}}',getTip:function(t,e){return t.renderTip(e.getAttribute(t.triangularSign))},getILeft:function(t,e){return t.left-21},getSHtml:function(t,e,n){return t.tpl.replace(/\{\{#baobiao_text\}\}/g,e)}}};function l(t){return this.type=t.type
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