interesting sequence n is a natural number and x1,x2,... is a sequence of numbers 1 and −1 with these properties: it is periodic and its least period number is 2n−1. (it means that for every natural number j we have xj+2n−1=xj and 2n−1 is the least number with this property.) There exist distinct integers 0≤t1<t2<...<tk<n such that for every natural number j we have xj+n=xj+t1×xj+t2×...×xj+tk Prove that for every natural number s that s<2n−1 we have 2n−1∑i=1xixi+s=−1 Time allowed for this question was 1 hours and 15 minutes.