prove that (n-2)(n-1)(2n-3) is divisible by 6 if n is any positive integer greater than 2

Answer:Proof is in the explanation.Step-by-step explanation:I'm going to use mathematical induction.That means we are going to show:1) For n=3 the expression given is a multiple of 6. (We started at n=3 because it says n>2.)2) If the base cases check out, then we are going to assume (n-2)(n-1)(2n-3) is a multiple of 6, then show ([n+1]-2)([n+1]-1)(2[n+1]-3) is also a multiple of 6.-----------------------------------------------------------------------------------------Proof:Base case (n=3):(3-2)(3-1)(2*3-3)1(2)(6-3)2(3)6 6 is a multiple of 6 since 6(1)=6.After the base case (for all natural numbers greater than 2):Assume there is integer k such that:6k=(n-2)(n-1)(2n-3).We are going to show 6m=([n+1]-2)([n+1]-1)(2[n+1]-3) where m is a integer. ([n+1]-2)([n+1]-1)(2[n+1]-3) (n-1)(n)(2n-1)(n-2+1)(n)(2n-1)(n-2)(n)(2n-1)+1(n)(2n-1)(n)(n-2)(2n-1)+1(n)(2n-1)(n-1+1)(n-2)(2n-1)+1(n)(2n-1)(n-1)(n-2)(2n-1)+1(n-2)(2n-1)+1(n)(2n-1)(2n-1)(n-2)(n-1)+1(n-2)(2n-1)+1(n)(2n-1)(2n-3+2)(n-2)(n-1)+1(n-2)(2n-1)+1(n)(2n-1)(2n-3)(n-2)(n-1)+2(n-2)(n-1)+1(n-2)(2n-1)+1(n)(2n-1)6k+2(n-2)(n-1)+1(n-2)(2n-1)+1(n)(2n-1)6k+2(n^2-3n+2)+1(2n^2-5n+2)+2n^2-n6k+6n^2-12n+66(k+n^2-2n+1)where k+n^2-2n+1 since integers are closed under addition and multiplication (referring to the n^2, the n*n part).Since we have found an integer m, k+n^2-2n+1, such that 6m=([n+1]-2)([n+1]-1)(2[n+1]-3) then we have shown for all integers greater than 2 we have that (n-2)(n-1)(2n-3) is divisible by 6.//