用裂项相消法解此题.拜托!已知an=1/[(n+1)(n-1)],求Sn.这题最后带着a1就行了.

an=1/[(n+1)(n-1)] = 2[1/(n-1) -1/(n+1)]
则a2=2(1/1 - 1/3)
a3=2(1/2 - 1/4)
a4=2(1/3 - 1/5)
a5=2(1/4 - 1/6)
...
a(n-1)=2[1/(n-2) - 1/n]
an = 2[1/(n-1) -1/(n+1)]
则Sn=a1/2+(1+1/2-1/n-1/(n+1)/2
=a1+(n²-3n-2)/n(n+1)

an=1/[(n+1)(n-1)]
=2/(n+1)(n-1)÷2
=[(n+1)-(n-1)]/(n+1)(n-1)÷2
=1/2(n-1)-1/2(n+1)
Sn=a1+a2+...+an
=a1+1/2-1/2(1+1)+...+1/2(n-1)-1/2(n+1)
=a1+1/2-1/2(n+1)