Question

已知数列｛an｝满足a1=2an+1=1+an/1-an则a1·a2·a3……·a2007=

Answer 1

a(n+1)=(1+an)/(1-an)
a(n+2)=[1+a(n+1)]/[1-a(n+1)]=[1+ (1+an)/(1-an)]/[1-(1+an)/(1-an)]=2/(-2an)=-1/an
a(n+4)=-1/a(n+2)=-1/(-1/an)=an，数列是以4为周期的周期数列。
a2=(1+a1)/(1-a1)=(1+2)/(1-2)=-3
a1·a2·a3·a4=(a1·a3)·(a2·a4)=[a1·(-1/a1)]·[a2·(-1/a2)]=1
即数列从第1项开始，每4项一组，各组的乘积都=1
a2008=a(4×501+4)=a4=-1/a2=-1/(-3)=1/3
a1·a2·a3·a4·...·a2007
=a1·a2·...·a2008/a2008
=a1·a2·a3·a4)·(a5·a6·a7·a8)·...·(a2005·a2006·a2007·a2008)/a2008
=1/a2008
=1/(1/3)
=3

Answer 2