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._数学解答

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.

人气：364 ℃　时间：2020-05-09 16:47:19

解答

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)