1/1*2=1/1-1/2,1/2*3=1/2-1/3,1/3*4=1/3-1/4,...

来源:百度知道 编辑:UC知道 时间:2024/09/24 11:26:40
1/1*2=1/1-1/2,1/2*3=1/2-1/3,1/3*4=1/3-1/4,...根据你发现的规律,计算:2/1*2+2/2*3+2/3*4+...+2/n*(n+1)=? (n为正整数)

2/1*2+2/2*3+2/3*4+...+2/n*(n+1)
=2(1/2-1/2*3+...+1/n(n+1)
=2(1-1/2+1/2-1/3+...+1/n-1/n+1)
=2(1-1/n+1)
=2n/n+1

2/1*2+2/2*3+2/3*4+...+2/n*(n+1)
=2(1-1/2+1/2-1/3+...+1/n-1/(n+1))
=2(1-1/(n+1))
=2n/(n+1)

∵1/[n*(n+1)]=[(n+1)-n]/[n*(n+1)]=1/n-1/(n+1)
∴2/(1*2)+2/(2*3)+2/(3*4)+...+2/[n*(n+1)]=2{1/(1*2)+1/(2*3)+1/(3*4)+...+1/[n*(n+1)]=2[1/1-1/2+1/2-1/3+1/3-1/4+...+1/n-1/(n+1)]=2[1-1/(n+1)]=2n/(n+1)

原式=2【(1)-(1/2)+(1/2)-(1/3)+(1/3)-(1/4)...(1/n)-(1/n+1)】
=2(1-1/n+1)
=2n/n+1

原式=2/1-2/2+2/2-2/3+2/3……+2/n-2/n+1=2-2/n+1=2n/(n+1)