SOLUTION: I was asked to do a proof by math induction for the formula of diagnols in a polygon. The formula for diagnols in a polygon is n(n-3)/2. N is the number of sides in any given pol

Question 124281: I was asked to do a proof by math induction for the formula of diagnols in a polygon. The formula for diagnols in a polygon is n(n-3)/2. N is the number of sides in any given polygon. The basis for the proof is n=3. For the basis step, I am asked to show how the generalization is true for n=3. I must show the induction hypothesis for n=k. To complete the proof by math induction, I must show that what is true for k is also true for k+1. I am asked to show the steps and leave it in simplified form. I am also asked to show the algebra that demonstrates how what is true for k is also true for k+1.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I was asked to do a proof by math induction for the formula of diagnols in a polygon. The formula for diagnols in a polygon is n(n-3)/2. N is the number of sides in any given polygon. The basis for the proof is n=3. For the basis step, I am asked to show how the generalization is true for n=3. I must show the induction hypothesis for n=k. To complete the proof by math induction, I must show that what is true for k is also true for k+1. I am asked to show the steps and leave it in simplified form. I am also asked to show the algebra that demonstrates how what is true for k is also true for k+1.
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If n=3 (triangle) there are zero diagonals:
N(3) = 3(3-3)/2 = 0
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Assume the formula is true for n=k
N(k)= (k)(k-3)/2
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Prove the formula is true for n=k+1
Comment: You must know that the number of diagonals increases by n-2
for each increase of one side. Try it for n=3,4,5 and you will see
what I mean.
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If n= k+1 the formula becomes:
N(k+1) = N(k)+[(k+1)-2]
N(k+1) = [k(k-3)/2] + [k-1]
N(k+1) = [k(k-3) + 2k-2]/2
N(k+1) = [k^2-3k+2k-2]/2
N(k+1) = [k^2-k-2]/2
N(k+1) = [(k+1)(k-2)]/2
N(k+1) = [(k+1)((k+1)-3)]/2
And that is what we were trying to prove.
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Cheers,
Stan H.
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