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Question 1: Prove that (9n+6)^2 -(9n-6)^2 is always a multiple of 9 |

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 4 consecutive numbers is always even |

Question 4: Prove the identity (n+2)^2 -(n+8) ≡ (n+4)(n-1) |

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Question 1: (9n+6)^2 -(9n-6)^2 =(81n^2+108n +36)-(81n^2-108n +36) =216n=9 \times 24n Therefore (9n+6)^2 -(9n-6)^2 is a multiple of 9 |

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)+(n+3) =4n+6 =2(2n+3) Any multiple of 2 is always even, therefore the sum of 4 consecutive numbers is always even |

Question 4: (n+2)^2 -(n+8) =(n^2+4n +4)-(1n+8) =n^2+3n-4 =(n+4)(n-1) |