周期性函数递推设数列｛an｝满足a1=a2=1,a3=2,且对任意正整数n都有an*an+1*an+2≠1,又an*an+1*an+2*an+3=an+an+1+an+2+an+3,求a1+a2+…a100

Question

周期性函数递推
设数列｛an｝满足a1=a2=1,a3=2,且对任意正整数n都有an*an+1*an+2≠1,又an*an+1*an+2*an+3=an+an+1+an+2+an+3,求a1+a2+…a100

Answer 1

an*a(n+1)*a(n+2)*a(n+3)=an+a(n+1)+a(n+2)+a(n+3)
a(n+1)*a(n+2)*a(n+3)*a(n+4)=a(n+1)+a(n+2)+a(n+3)+a(n+4)
a(n+4)/an=[an+a(n+1)+a(n+2)+a(n+3)]/[an+a(n+1)+a(n+2)+a(n+3)]
[a(n+4)-an][a(n+1)+a(n+2)+a(n+3)]=0
∵a1+a2+a3=4
∴a(n+1)+a(n+2)+a(n+3)≠0
a(n+4)=an
1*1*2*a4=1+1+2+a4
a4=4
a5=a1,a6=a2,a7=a3a8=a4,...
a1+a2+…a100=25*(a1+a2+a3+a4)=200