已知数列{An},且An=1⼀n(n+2)，求Sn

An=1/n(n+2)

=0.5*[1/n-1/(n+2)]

所以，Sn=A1+A2+...+A(n-1)+An

=0.5*[1/1-1/3+1/2-1/4+1/3-1/5......+1/(n-3)-1/(n-1)+1/(n-2)-1/n]
=0.5*[1.5-1/(n-1)-1/n]

=0.75-1/[2n(n-1)]

这是数列求和的常用方法：裂项相消

an=1/n(n+2)
a1=1/1*3=1/2(1-1/3)
a2=1/2*4=1/2(1/2-1/4)
a3=1/3*5=1/2(1/3-1/5)
...
an-1=1/(n-1)(n+1)=1/2(1/(n-1)-1/(n+1))
an=1/n(n+2)=1/2(1/n-1/(n+2))
相加,有:
Sn=1/2(1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+...+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/n(n+2)=1/2（1/N-1/(N+2)）
则Sn＝1/2（1－1/3+1/2-1/4+1/3-1/5+......1/（n－1）－1/（n＋1）＋1/n－1/（n＋2））＝1/2(1+1/2-1/(n＋1)－1/（n＋2）)
＝1/2(3/2-1/(n＋1)－1/（n＋2）)
下面就不化简了，谢谢

2An=2/(n)(n+2)=1/n-1/(n+2)
2Sn=1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7……
可以抵消
n为奇数时……
2Sn＝3/2-1/(n+1)-1/(n+2)
Sn=3/4-1/[2(n+1)]-1/[2(n+2)]
n为偶数时……
2Sn=3/2-1/(n+1)-1/(n+2)
Sn=3/4-1/[2(n+1)]-1/[2(n+2)]
所以

Sn=3/4-1/[2(n+1)]-1/[2(n+2)]