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）
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）
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)]
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
已知数列{an}的通项公式是an=1\/{n(n+2) }(n∈N),求它的前n项的和。
an=1\/n(n+2)={1\/n-1\/(n+2)}\/2 a1+a2+...+an=(1\/2)*{1-1\/3+1\/2-1\/4+1\/3-1\/5+1\/4-1\/6+...+1\/n-1\/(n+2)}={1+1\/2-1\/n-1\/(n+2)}\/2
已知数列an=1\/n(n+1)(n+2),求数列的前n项和Sn 最好利用裂项法
an=1\/2*2\/[n(n+1)(n+2)]=1\/2*[(n+2)-n]\/[n(n+1)(n+2)]=1\/2{[(n+2)\/[n(n+1)(n+2)]-n\/[n(n+1)(n+2)]=1\/2{1\/[n(n+1)]-1\/[(n+1)(n+2)]} 所以Sn=1\/2*{1\/1*2-1\/2*3+1\/2*3-1\/3*4+……+1\/[n(n+1)]-1\/[(n+1)(n+2)]} =1\/...
已知an=1\/(n+1)(n+2),求an的前n项和sn
an=1\/(n＋1)(n＋2)a1=1\/(1＋1)(1＋2)=1\/2-1\/3 an的前n项和sn=[1\/(1+1)-1\/(1+2)]+[1\/(2+1)-1\/(2+2)]+……+[1\/(n+1)-1\/(n+2)]=1\/(1+1)-1\/(n+2)=n\/(2n+4)
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
已知数列{an}的通项公式为an=n+1\/2的(n+1)次方,求数列前n项和sn
解：an=n+1\/2^(n+1)，则 Sn=a1+a2+.+an =(1+2+.+n)+(1\/2^2+1\/2^3+.+1\/2^(n+1)) （分别是等差数列和等比数列）=(n+1)n\/2+1\/2^2(1-1\/2^n)\/(1-1\/2)=(n+1)n\/2+1\/2-1\/2^(n+1)。
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...
