Question

已知an＝1/(n+3)n，求sn

Answer 1

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)]

Answer 2

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)