Respuesta :
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.
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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-n
6k+6n^2-12n+6
6(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.
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