来自饶建辉的问题
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)(2n+1)的值为17/35,求n的值
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)(2n+1)的值为17/35,求n的值
1回答
2020-10-1004:22
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)(2n+1)的值为17/35,求n的值
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)(2n+1)的值为17/35,求n的值
*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