已知an＝1/(n+3)n，求sn
答案:2 悬赏:10
解决时间 2021-01-28 13:41
- 提问者网友：蓝琪梦莎
- 2021-01-27 15:43
已知an＝1/(n+3)n，求sn
最佳答案
- 二级知识专家网友：三千妖杀
- 2021-01-27 17:11
an=1/[n(n+3)]=(1/3)[1/n -1/(n+3)]
Sn=a1+a2+...+an
=[1/1-1/4+1/2-1/5+...+1/n -1/(n+3)]
=(1/1+1/2+...+1/n)-[1/4+1/5+...+1/n+1/(n+1)+1/(n+2)+1/(n+3)]
=1/1+1/2+1/3-1/(n+1)-1/(n+2)-1/(n+3)
=11/6 -1/(n+1)-1/(n+2)-1/(n+3)
Sn=a1+a2+...+an
=[1/1-1/4+1/2-1/5+...+1/n -1/(n+3)]
=(1/1+1/2+...+1/n)-[1/4+1/5+...+1/n+1/(n+1)+1/(n+2)+1/(n+3)]
=1/1+1/2+1/3-1/(n+1)-1/(n+2)-1/(n+3)
=11/6 -1/(n+1)-1/(n+2)-1/(n+3)
全部回答
- 1楼网友：胯下狙击手
- 2021-01-27 18:02
an=1/(n+3)n=1/3[1/n-1/(n+3)]
Sn=a1+a2+a3+...+an
=1/3[1/1-1/4
+1/2-1/5
+1/3-1/6
+1/4-1/7
+...
+1/(n-3)-1/n
+1/(n-2)-1/(n+1)
+1/(n-1)-1/(n+2)
+1/n-1/(n+3)]
=1/3[1/1+1/2+1/3-1/(n+1)-1/(n+2)-1/(n+3)]
=11/18-1/3[1/(n+1)+1/(n+2)+1/(n+3)]
Sn=a1+a2+a3+...+an
=1/3[1/1-1/4
+1/2-1/5
+1/3-1/6
+1/4-1/7
+...
+1/(n-3)-1/n
+1/(n-2)-1/(n+1)
+1/(n-1)-1/(n+2)
+1/n-1/(n+3)]
=1/3[1/1+1/2+1/3-1/(n+1)-1/(n+2)-1/(n+3)]
=11/18-1/3[1/(n+1)+1/(n+2)+1/(n+3)]
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