Question

запишите в виде дроби n+2/n! -3n+2/(n+1)!

Answer 1

(n+2)/n1-(3n+2)/[n!(n+1)]=[(n+1)(n+2)-(3n+2)]/(n+1)!=
=(n²+2n+n+2-3n-2)/(n+1)!=n²/(n+1)!