1,1/(1+2),1/(1+2+3),…1/(1+2+3+…+n),…,求它的前n项和
来源:百度知道 编辑:UC知道 时间:2024/06/29 01:38:04
哈哈
9
an=2/(n*(n+1))
Sn=2*(1/1-1/2+1/2-1/3+...+1/(n-1)-1/n+1/n-1/(n+1))
Sn=2*(1-1/(n+1))
Sn=2*n/(n+1)
它的前n项和
1+1/3+1/6+1/10+…+1/[n(n+1)/2]=2/2+2/6+2/12+2/20+…+2/[n(n+1)]
=2{1/1*2+1/2*3+1/3*4+1/4*5+…+1/[n(n+1)]}=2[(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+…+(1/n-1/n+1)]=2[1-1/n+1]=2n/(n+1)
解:
将已经给数列分母求和可以知道,通项公式为:
1/(n*(n+1)/2)=
=2/(n*(n+1)=2(1/n-1/(n+1))
所以原来的数列的和等价于
2*[(1-1/2)+(1/2-1/3),....+(1/n-1/(n+1))]=
2*[1+(-1/2+1/2)+(-1/3+1/3)...+(-1/n+1/n)-/(n+1)]=
2*[1-1/(n+1)]=2n/(n+1)
1+2+3+…+n=(1+n)n/2
倒数为2(1/n-1/(n+1))
所以前n项和为2-1/n
an=2/(n*(n+1))
Sn=2*(1/1-1/2+1/2-1/3+...+1/(n-1)-1/n+1/n-1/(n+1))
Sn=2*n/(n+1)
(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1)
1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+100)
1+1/(1+2)+1/(1+2+3)+-------+1/(1+2+3+----+100)
1+1/1+2+1/1+2+3+...+1/1+2+3...+2000
1+1/1+2+1/1+2+3.........+1/1+2+3.....100
1*(1/1+2)*(1/1+2+3)*~~~*(1/1+2+~~~2005)=?
(1-1/2)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000)
1+1/2+1+1/3+1+1/4+......+1/100=?
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)
(1-1/2^2)*(1-1/3^2)*(1-1/4^2).......(1-1/100^2)