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Algebra/algebra , vector question, age 17, 11th grade

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Question
1.The position vectors of vertices A, B and C of  triangle ABC are a b

, and c

respectively.
Find the position vector of the centroid of triangle ABC.

2prove that
(2n)!/n!= 1*3*5.....(2n-1)*2^n

Answer
The centroid of a triangle is the intersection of the bisecting lines of each angle.
To get the actual position in this way, calculate the midpoint of one side and then
determine the line that goes through this point by using that point and the opposite
corner of the triangle.  Do this to get a line for each side.  The centroid is the
intersection of these three lines.  Only two lines are needed, and the third line
is used to check and male sure it intersects there as well.

Look at when n-3.  That is where 2n=6.
It is known that 6! is 1*2*3*4*5*6 and 3! is 1*2*3.
If each of the number in the 3! is multiplied by 2, we get 2*4*6/2^3.
This gives 6!/3! = (1*2*3*4*5*6)/(2*4*6/2^3).
Since the bottom cancels the even terms, and the bottom is divided by 2^3,
this is the same as 1*3*5*2^3.

For each increase in n, that multiplies by one more odd number and one more even number to (2n)!
The numbers added are (2n-1) and 2n.  Since we have an n added down below, cancelling the n's
gives (2n-1)2.  That is where we get the formula.

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