an=2⼀（n(n+2))求和

用列项法
an=2/（n(n+2))=1/n-1/(n+2)
然后楼主自己写出前几项就会发现前后两项相互约去
最后的结果是3/2-（2n+3)/(n+1).(n+2)

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

所以Sn=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+1/2-1/(n+1)-1/(n+2)
=(3n²+7n+3)/(2n²+6n+4)

an=2/（n(n+2))=1/n-1/(n+2)
∴Sn=1/1-1/3+1/2-1/4+1/3-1/5+……+1/n-1/(n+2)
=3/2-1/（n+1）-1/（n+2）=（3n²+9n)/2(n+1)(n+2)

化简成an＝1/n-1/(n+2),求和从a1加到an就是1＋1/2-1/(n+1)-1/(n+2)

(n+1)-1/an=[(n+2)-n]/n(n+2)
=(n+2)/n(n+2)-n/2-1/(n+2)
=(3n²+7n+3)/n(n+2)
=1/n-1/(n+2)
所以Sn=1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+……+1/(n-1)-1/(n+1)+1/(n+2)
=1+1/n-1/

1/2