Question

(n+2)!=132*n!
комбинаторика

Answer 1

$(n+2)!=132*n!\ frac{ (n+2)!}{n!}=132\ (n+1)(n+2)=132\ n^2+3n+2=132\ n^2+3n-130=0\ (n+13)(n-10)=0\ n=10$