{an}的前10项之和=(1/2)-(1/3)+(1/3)-(1/4)+(1/4)-(1/5)+...+(1/11)-(1/12)
=(1/2)-(1/12)
=5/12本回答被提问者采纳
数列{an}的通项公式为an=1\/(n+1)(n+2),则{an}的前10项之和为
an=1\/(n+1)(n+2)=(1\/(n+1))-(1\/(n+2)){an}的前10项之和=(1\/2)-(1\/3)+(1\/3)-(1\/4)+(1\/4)-(1\/5)+...+(1\/11)-(1\/12)=(1\/2)-(1\/12)=5\/12
数列an的通项公式an=1\/(n+1)(n+2),则an的前10项之和为?
an=1\/(n+1)(n+2)=[1\/(n+1)]-[1\/(n+2)]Sn=1\/2-1\/3+1\/3-1\/4+···+[1\/(n+1)]-[1\/(n+2)]=1\/2-[1\/(n+2)]将n=10带入 答案为5\/12
数列an的通项公式是an等于n乘以n加1,若前n项的和为十一分之十,则项数...
回答：解答: 题目的通项公式应该是an=1\/[n*(n+1)]=1\/n-1\/(n+1) ∴ 前n项和=1\/1-1\/2+1\/2-1\/3+1\/3-1\/4+.....+1\/n-1\/(n+1)=1-1\/(n+1)=n\/(n+1)=10\/11 ∴ n=10 即项数是10
已知数列{an}的通项公式为an=1\/(n+2)(n+1),则前100项和为?
an=[(n+2)-(n-1)]\/(n+1)(n+2)=(n+1)\/(n+1)(n+2)-(n+1)\/(n+1)(n+2)=1\/(n+1)-1\/(n+2)所以S100=1\/2-1\/3+1\/3-1\/4+……+1\/101-1\/102 =1\/2-1\/102 =25\/51
数列{an}的通项公式为an=1(n+1)(n+2),则该数列的前n项和Sn=___
an＝1(n+1)(n+2)=1n+1?1n+2∴Sn=(12?13)+(13?14)+(14?15)+…+（1n+1?1n+2）=12?1n+2=n2(n+2)．故答案为n2(n+2)
已知数列an的通项公式an=1\/(n+2)(n+1),则其前n项和Sn
an=1\/(n+2)(n+1)=1\/(n+1）-1\/（n+2）所以sn=1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+.1\/n-1\/（n+1）+1\/(n+1）-1\/（n+2）=1\/2-1\/（n+2）=n\/(2n+4）
数列{an}的通项公式为an=1\/(2n+1)^2,求前n项和Sn
【答案】：style='color:#fe2419;'>an=2n+1(当n为奇数)style='color:#fe2419;'>an=2∧n (当n为偶数)style='color:#fe2419;'>要求n项的和必须n为连续自然数;所以 设奇数n=2m-1;偶数n=2m.（m为连续自然数）style='color:#fe2419;'>所以1、奇数项:an=2n+1=2(2m-1)+1=4m-1...
已知数列an的通项公式为an=1\/(n(n+1)(n+2)),求数列an的前n项和Sn
an=1\/2*[1\/n - 2\/(n+1) +1\/(n+2)]Sn=1\/2{(1\/1 -2\/2 + 1\/3)+(1\/2 - 2\/3 +1\/4)+...+ [1\/n - 2\/(n+1) +1\/(n+2)]} =1\/2[1\/1 -1\/2 - 1\/(n+1) +1\/(n+2)]=1\/4-1\/[2*(n+1)(n+2)]
数列{an}的通项为an=1\/2n(2n+2),求{an}的前n项和
an=1\/[2n(2n+2)]= (1\/4) [ 1\/n - 1\/(n+1) ]Sn =a1+a2+...+an =(1\/4) [1-1\/(n+1) ]= n\/[4(n+1)]
若数列an的通项为1\/n(n+1)(n+2),求an的前n项和
简单计算一下即可，详情如图所示
