1/1x2+1/2x3+...1/n(n-1)=98/99,n的值是多少
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 10:21:18
1/1x2+1/2x3+...1/n(n-1)=98/99,n的值是多少
1/1x2+1/2x3+...1/n(n-1)=98/99,n的值是多少
1/1x2+1/2x3+...1/n(n-1)=98/99,n的值是多少
1/1×2+1/2×3+...+1/n(n-1)
=(1-1/2)+(1/2-1/3)+...+[1/(n-1)-1/n]
=1-1/n
1-1/n=98/99=1-1/99
所以n=99
1/1x2+1/2x3+...1/n(n-1)
=(1/1-1/2)+(1/2-1/3)+……+(1/n-1/n-1)
=1-1/n-1
而98/99=1-1/99=1-1/100-1
所以n的值是100
1/n(n-1)=1/(n-1)-1/n
原式=1-1/2+1/2-1/3+1/3-……+1/(n-1)-1/n
=(n-1)/n
n=99
1x2+2x3+3x4+4x5+.+n(n+1)+n(n+2)=?
1X2+2X3+3X4+.NX(N+1)
1X2+2X3+3X4+.NX(N+1)规律
1X2+2X3.+nx(n+1)
1 1X2+2X3+3X4.n(n+1)(n+2)/
1x2+2x3+3x4+.+n(n+1)=_____(n为自然数)
1/1X2+1/2X3+1/3X4+...+1/n(n+1)等于几?
1/1x2+1/2x3+1/3x4+...+1/n(n+1)怎么算
计算:1/1x2+1/2x3+1/3x4+.+1/(n-1)n
求Sn=1/1x2+1/2x3+1/3x4+.+1/n(n+1)
设X1~N(1,2),X2~N(0,3),X3~N(2,1),X1,X2,X3相互独立,则P{0
若1/1x2+1/2x3+.+1/nx(n+1)=98/99求n
证明:1/√(1x2)+1/√(2x3)+……+1/√(n(n+1))
1X2分之1+2x3分之1+.+n(n+1)分之一
1x2+2x3+3x4+4x5.n(n+1)等于多少?急救!
求使1x2+2x3+3x4+…+n(n+1)
1x2+2x3+3x4+...+n(n+1)=?
1x2+2x3+3x4...+10x11=?1x2+2x3+3x4...+nx(n-1)=?