所以对式子裂项相加
Sn=2/2+2/(2*3)+...+2/(n*(n+1))
把2提出来
Sn=2(1/2+1/(2*3)+....+1/(n*(n+1))
Sn=2(1-1/2+1/2-1/3+......+1/n-1/(n+1))
Sn=2(1-1/(n+1))
Sn=2-2/(n+1)
求Sn=1+1\/(1+2)+1\/(1+2+3)+...+1\/(1+2+3+...+n)
因为1\/(1+2+3...+n)=2\/(n*(n+1))所以对式子裂项相加 Sn=2\/2+2\/(2*3)+...+2\/(n*(n+1))把2提出来 Sn=2(1\/2+1\/(2*3)+...+1\/(n*(n+1))Sn=2(1-1\/2+1\/2-1\/3+...+1\/n-1\/(n+1))Sn=2(1-1\/(n+1))Sn=2-2\/(n+1)
数列求和:Sn=1+1\/(1+2)+1\/(1+2+3)+...+1\/(1+2+3+4+...+n)
分母的通项是an=1+2+...+n=n(n+1)\/2 所以Sn=1\/a1+1\/a2+...+1\/an =2\/1*2+2\/2*3+...+2\/n(n+1)=2[(1-1\/2)+(1\/2-1\/3)+...+(1\/n-1\/(n+1))]=2[1-1\/(n+1)]=2n\/(n+1)
计算sn=1+1\/(1+2)+1\/(1+2+3)+...1\/(1+2+3+...+n)=
因为1\/(1+2+3...+n)=2\/(n*(n+1))所以对式子裂项相加 Sn=2／2+2／(2*3)+...+2\/(n*(n+1))把2提出来 Sn=2（1／2+1／(2*3)+...+1\/(n*(n+1)）Sn=2(1-1\/2+1\/2-1\/3+...+1\/n-1\/(n+1))Sn=2(1-1\/(n+1))Sn=2-2\/(n+1)
猜想sn=1+1\/(1+2)+1\/(1+2+3)+…+1\/(1+2+3+…+n)的表达式,并用数学归 ...
+1)]由于 A = 2 企鹅 A + B = 0 所以 B = -2 即1\/Tn = 2[1\/n - 1\/(n + 1)]Sn = 1\/T1 + 1\/T2 + 1\/T3 + ...+ 1\/Tn = 2[1\/1 - 1\/(1 + 1)+ 1\/2 -
用裂项相消法sn=1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+n)
用裂项相消法sn=1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+n) 我来答 你的回答被采纳后将获得: 系统奖励15(财富值+成长值)+难题奖励30(财富值+成长值)1个回答 #热议# 已婚女性就应该承担家里大部分家务吗?wangzunjing111 2015-03-26 · TA获得超过398个赞 ...
1+1\/(1+2)+1\/(1+2+3)+…+1\/(1+2+3+…+n)这个怎么计算啊
所以Sn=1+[1\/(1+2)]+〔1\/(1+2+3)〕+[1\/(1+2+3+4)]+……+[1\/(1+2+3+……+n)]=2[1\/1-1\/2]+2[1\/2-1\/3]+2[1\/3-1\/4]+...+2[1\/n-1\/(n+1)]=2[1-1\/2+1\/2-1\/3+1\/3-1\/4+...+1\/(n-1)-1\/n+1\/n-1\/(n+1)]=2[1-1\/(n+1)]=2n\/(n+1...
1+1\/(1+2)+1\/(1+2+3)+...+1\/(1+2+3+...+n)=?
解：等差数列1+2+。。。+n有n项，各项的平均值为(n+1)\/2，所以1+2+3+。。。+n=n*[(n+1)\/2]，所以 1\/(1+2+3+...+n)=2\/(n*(n+1))。
1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...1\/(1+2+3+...+100)=?
解：1+1\/（1+2）+1\/（1+2+3）+1\/（1+2+3+4)+...1\/(1+2+3+...+100)=2\/(1×2)+2\/(2×3)+2\/(3×4)+...+2\/(100×101)=2×[(1-1\/2)+(1\/2-1\/3)+...+(1\/100-1\/101)]=2×(1-1\/101)=2×100\/101 =200\/101 类似：1\/[a*(a+n)]=(1\/n)*[1\/a-...
1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+4+...+10)=?
(问题可如下分析:通项公式为an=2\/[(1+n)*n],求sn,其中*代表乘号.该类问题可将an裂项为2\/[1\/n-1\/(n+1)],然后相加即可)1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+4+...+10)=2\/[(1+1)*1]+2\/[(1+2)*2]+2\/[(1+3)*3]+2\/[(1+4)*4]+......
...+ 1\/1+2+3 + 1\/1+2+3+4 + ... + 1\/1+2+3+...+2011+2012 + 1\/1+2...
分母是等差数列和= (1+n)*n\/2 首项加末项乘以项数除以2 由此可知，sn= 2\/(2*1) + 2\/(3*2) + 2\/(4*3) + 2\/(5*4)…… + 2\/(2014*2013) + 2\/(2015*2014)把分子的2提取出来，再把每项的乘拆开 sn\/2 = (-1\/2 + 1\/1) + (-1\/3 + 1\/2) ...
