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

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-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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f(x)=2/3x+4/3 is the answer because is whose power in the senate is equal to the power of the speaker of the house?

step-by-step explanation:

step-by-step explanation:

two pairs of adjacent sides are of equal length. this implies that one diagonal divides the kite into congruent triangles, and so the angles between the two pairs of equal sides are equal in measure. it also implies that the diagonals are perpendicular.