Name: Date:

Question 1: Prove the identity (n-3)^2 -(6n-27) ≡ (n-6)(n-6) |

Question 2: Prove that the difference of the squares of 2 consecutive numbers is always the sum of the 2 numbers |

Question 3: Prove that the sum of 3 consecutive numbers is always a multiple of 3 |

Question 4: Prove that (6n+9)^2 -(6n-9)^2 is always a multiple of 9 |

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# Answers

**Question 1:**

(n-3)^2 -(6n-27)

=(n^2-6n +9)-(6n-27)

=n^2-12n+36

=(n-6)(n-6)

**Question 2:**

(2n+1)^2-(2n)^2

=4n^2+4n+1-4n^2

=4n+1=(2n+1)+(2n)

therefore the difference of the squares of 2 consecutive numbers is always the sum of the 2 numbers

**Question 3:**

n+(n+1)+(n+2)

=3n+3

=3(n+1)

Therefore the sum of 3 consecutive numbers is always a multiple of 3

**Question 4:**

(6n+9)^2 -(6n-9)^2

=(36n^2+108n +81)-(36n^2-108n +81)

=216n=9 \times 24n

Therefore (6n+9)^2 -(6n-9)^2 is a multiple of 9

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