求数列｛1/n(n+1)(n+2)｝的前n项和Sn

Question

求数列｛1/n(n+1)(n+2)｝的前n项和Sn 怎么计算

Answer 1

通项公式为An＝1/n(n+1)(n+2)
＝[1/n(n+1)-1/(n+1)(n+2)]/2
所以Sn＝[1/1*2－1/2*3]/2+
[1/2*3－1/3*4]/2+[1/3*4－1/4*5]/2+
……+[1/(n-1)*n－1/n*(n+1)]/2+
[1/n(n+1)-1/(n+1)(n+2)]/2
＝[1/1*2-1/(n+1)(n+2)]/2
＝1/4-1/[2(n+1)(n+2)]

Answer 2

1/n(n+1)(n+2)=0.5*(n+2-n)/n(n+1)(n+2)=0.5*[1/n(n+1)-1/(n+1)(n+2)]=0.5*[1/n-1/(n+1)-1/(n+1)+1/(n+2)]
就这的往下写，中间就消掉好多项了。裂项相消的思想。

Answer 3

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

所以Sn=(1/2)*{(1/1*2-1/2*3)+(1/2*3-1/3*4)+……+[1/n(n+1)-1/(n+1)(n+2)]}
=(1/2)[1/1*2-1/(n+1)(n+2)]
=(n^2+3n)/[4(n+1)(n+2)]

Answer 4

1/(1×2×3)+1/(2×3×4)+1/(3×4×5)……1/[n×(n+1)×(n+2)
=(1-1/(n+2)
=(n+1)/(n+2)