*3+1/3*4+...+1/n(n+1)
=1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)
1/1*3+1/3*5+1/5*7+.1/(2n-1)(2n+1)=17/35
1/2*[1-1/3+1/3-1/5+1/5-1/7……+1/(2n-1)-1/(2n+1)]=17/35
1-1/(2n+1)=34/35
1/(2n+1)=1/35
2n+1=35
n=17
探究1/1*2+1/2*3+1/3*4+...+1/n(n+1)= 若1/1*3+1/3*5+1/5*7+.1/(2n-
1个回答
相关问题
-
1.1/1*2+1/2*3+1/3*4+……+1/n(n+1) 2.1/1*3+1/3*5+1/5*7+……+1/(2n
-
1/1*3+1/3*5+1/5*7+.+1/（2n-1）(2n+1)=n/(2n+1)
-
1/(√3 +1) +( 1/√5+√3 )+ (1/√7+√5)+.+{1/√(2n+1)+√(2n-1)}
-
1*2+2*3+3*4+4*5+5*6+6*7+…+n(n+1)=n(n+1)(n+2)/3
-
1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+.+1/[n(n+)(n+)]
-
1/1*5+1/3*7+1/5*9+.+1/(2n-1)(2n+3)
-
若1+3+5……+（2n+1）=110[1/(1*2)+ 1/(2*3)+……+1/(n*(n+1))],则n=
-
数学归纳法证明-1+3-5+.+(-1)^n(2n-1)=(-1)^n*n1/1*3+1/3*5+1/5*7+.+1/(
-
求和1/1·3+1/3·5+1/5·7+...+1/(2n+1)(2n-1)
-
求和:1/(1*3)+1/(3*5)+1/(5*7)+...+1/(2n-1)(2n+1)=?