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Question 1: Prove that (5n+7)^2 -(5n-7)^2 is always a multiple of 5 |

Question 2: Prove that the sum of 3 consecutive numbers is always a multiple of 3 |

Question 3: Prove that the difference of the squares of 2 consecutive numbers is always the sum of the 2 numbers |

Question 4: Prove the identity (n+5)^2 -(5n+75) ≡ (n+10)(n-5) |

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Question 1: (5n+7)^2 -(5n-7)^2 =(25n^2+70n +49)-(25n^2-70n +49) =140n=5 \times 28n Therefore (5n+7)^2 -(5n-7)^2 is a multiple of 5 |

Question 2: n+(n+1)+(n+2) =3n+3 =3(n+1) Therefore the sum of 3 consecutive numbers is always a multiple of 3 |

Question 3: (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 4: (n+5)^2 -(5n+75) =(n^2+10n +25)-(5n+75) =n^2+5n-50 =(n+10)(n-5) |