求数列{1⼀n(n+2)}前n项和

1/n-1/(n+2)=(n+2)/n(n+2) - n/n(n+2) = 2/n(n+2)
2An=1/n - 1/(n+2)
2Sn=1/1 - 1/3 + 1/2 -1/4 + 1/3 - 1/5 + 1/4 - 1/6 + ... + 1/(n-3) - 1/(n-1) + 1/(n-2) - 1/n+ 1/(n-1) - 1/(n+1) + 1/n - 1/(n+2)
=1+1/2-1/(n+1) - 1/(n+2)
=3/2 - (2n+3)/[(n+1)(n+2)]
Sn=3/4 - 1/[2(n+1)(n+2)]

1/[n(n+2)]=(1/2)[1/n -1/(n+2)]
Tn=(1/2)[1/1-1/3+1/2-1/4+...+1/n-1/(n+2)]
=(1/2)[(1/1+1/2+...+1/n)-(1/3+1/4+...+1/(n+2))]
=(1/2)[1/1+1/2 -1/(n+1)-1/(n+2)]
=(3/4) -1/[2(n+1)] -1/[2(n+2)]

解:n=1／n(n +2)=[1／n-1／(n+2)]／2
Sn=[1／1-1／3+1／2-1／4+1／3-1／5+、、、+1／(n-2)-1／n+1／(n-1)-1／(n+1)+1／n-1／(n+2)]／2
=3／4-1／2(n+2)
S1=1／3
S2=1／3+1／8。

1/n(n 2)=1/n-1/(n 2)
前n项和=1＋1／2－1／（n 1)-1/(n 2)=3/2-1/(n 1)-1/(n 2)