Question

如何求1/n(n+2)的前n项和?

Answer 1

1/n(n+2)=1/2*(1/n-1/(n+2))
前n项和
=1/2*(1/1-1/3+1/2-1/4+1/3-1/5+……+1/(n-3)-1/(n-1)+1/(n-2)-1/n+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))

Answer 2

1/n(n+2) = 1/2 [1/n - 1/(n+2) ]
sN = 1/2 { [1/1-1/3]+[1/2-1/4]+[1/3-1/5]+...+[1/n-1/(n+2)]}
=1/2 { [1+1/2+1/3+1/4+...+1/n] - [1/3+1/4+1/5+...+1/(n+2)]}
=1/2 [1+1/2 -1/(n+1) - 1/(n+2)]

Answer 3

只求极限吗 极限n->无穷大时 极限为
3/2

Answer 4

1/n(n+2)=1/2[1/n-1/(n+2)]
提出1/2，然后相加就OK啦，你懂得