a1=1-1/2
a2=1/2-1/3
a3=1/3-1/4
..
..
..
an=1/n-1/(n+1)
sn= 1-1/2 + 1/2-1/3 + 1/3-1/4 + ... + 1/n-1/(n+1)
= 1 - 1/(n+1)
= n/(n+1)
sn= 1-1/2 + 1/2-1/3 + 1/3-1/4 + ... + 1/n-1/(n+1)
= 1 - 1/(n+1)
= n/(n+1)
Sn=1-1/(n+1)=n/(n+1)
求数列{1\/n(n+1) }的前n项和Sn
an=1\/n(n+1)=1\/n-1\/(n+1)Sn=a1+a2+……+an =(1-1\/2)+(1\/2-1\/3)+……+[1\/n-1\/(n+1)]=1+(1\/2-1\/2)+(1\/3-1\/3)+……+(1\/n-1\/n)-1\/(n+1)=1-1\/(n+1)=n\/(n+1)
求数列1\/n(n+1)的前n项和SN
1\/n(n+1)的前n项和SN an=1\/n(n+1)=1\/n-1\/n+1 an-1=1\/(n-1)-1\/n ...a2=1\/2-1\/3 a1=1-1\/2 所以Sn=1-1\/2+1\/2-1\/3+1\/3-1\/4...+1\/n-1\/n+1=1-1\/n+1=n\/(n+1)如有不明白，可以追问
求数列1\/n(n+1)的前n项和sn=1\/1*2+1\/2*3...1\/n*(n-1)!!!
1\/n(n+1）前n项和 因为1\/n(n+1）=1\/n - 1\/（n+1)所以sn=1-1\/2+1\/2-1\/3+1\/3-1\/4+...+1\/n - 1\/（n+1）=1-1\/（n+1）1\/n*(n-1) 同样的方法 1\/n*(n-1) =1\/（n-1) -1\/n （n大于等于2）所以sn=1-1\/2+1\/2-1\/3+1\/3-1\/4 ...
求数列{1\/n(n+1)}的前n项和
求数列的前n项和
求数列{1\/n(n+1]}的前n项和Sn 怎样推导
这种类型的式子，是显然应用裂项法处理的应用，如下：因为1\/n(n+1)=1\/n-1\/(n+1)，所以：Sn=1\/1*2+1\/2*3+...+1\/(n-1)n+1\/n(n+1)=(1-1\/2)+(1\/2-1\/3)+...+(1\/(n-1)-1\/n)+(1\/n-1\/(n+1))=1-1\/(n+1)=n\/(n+1).
数列{1\/n(n+1)}的前n项和Sn=1\/1*2+1\/2*3+1\/3*4+1\/4*5+...+1\/n(n+1...
解析，an=1\/{n(n+1)}=1\/n-1\/(n+1)那么，Sn=a1+a2+a3+……+an=1-1\/2+1\/2-1\/3+……+1\/n-1\/(n+1)=1-1\/(n+1)=n\/(n+1)推广，an=1\/{(n+t)(n+t+1)}(t∈自然数N)，都可以，这样拆开，an=1\/(n+t)-1\/(n+t+1)另外，an=1\/n(n-1)=1\/(n-1)-1\/n(n...
1\/ n(n+1)的前n项和怎么算?
1\/n(n+1)=1\/n-1\/(n+1)1\/(2n-1)(2n+1)=1\/2[1\/(2n-1)-1\/(2n+1)]1\/n(n+1)(n+2)=1\/2[1\/n(n+1)-1\/(n+1)(n+2)]1\/(√a+√b)=[1\/(a-b)](√a-√b)n·n!=(n+1)!-n!
数列{1\/n(n+1)}前n项求和公式及证明方法,谢谢
1\/n(n+1)=1\/n-1\/(n+1)所以前n项和为 1-1\/2+1\/2-1\/3+1\/3+...+1\/n-1\/(1+n)=1-1\/(1+n)
一个数列的通项公式为:an=1\/n(n+1),求前n项和Sn
裂项求和法：an =1\/n - 1\/(n+1),所以 Sn=1 - 1\/2 + 1\/2 - 1\/3 + … + 1\/（n-1) - 1\/n + 1\/n - 1\/(n+1)=1- 1\/(n+1)=n\/(n+1).中间都消掉了，比如第二项-1\/2和第三项1\/2。如果还有问题可以问我。
