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)
两式相减,整理有
(a(n+4)-an)(a(n+1)*a(n+2)*a(n+3)-1)=0
由a1=1,a2=1,a3=2,求出a4=4,a2*a3*a4=8不等于1,所以a(n+1)*a(n+2)*a(n+3)-1不恒等于0
所以
a(n+4)-an=0
所以数列是a1=1,a2=1,a3=2,a4=4的循环数列
循环数列 1,1,2,4,1,1,2,4...
S100=(1+1+2+4)*25=200