数列求和1/1*3,1/2*4,1/3*5...1/n(n+2)

来源：百度知道 编辑：UC知道 时间：2024/09/28 17:49:03
RT 具体过程

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

an
=1/(2n-1)(2n+1)
=1/2[1/(2n-1)-1/(2n+1)]

Sn
=1/2[1-1/3]+1/2[1/3-1/5]+1/2[1/5-1/7]+...+1/2[1/(2n-3)-1/(2n-1)]+1/2[1/(2n-1)-1/(2n+1)]
=1/2[1-1/3+1/3-1/5+1/5-1/7+...+1/(2n-3)-1/(2n-1)+1/(2n-1)-1/(2n+1)]
=1/2[1-1/(2n+1)]
=n/(2n+1)

1/n(n+2)=(1/2)[1/n-1/(n+2)] 原式=[1+1/2-1/(n+1)-1/(n+2)]/2

1/n(n+2)=(1/n - 1/(n+2))/2