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Objective: using algebra to prove number facts 1. Prove that the square of any odd number is always one more than a multiple of 8. 2. Show that the sum of any three consecutive multiples of 3 is also a multiple of 3. 3. Write down the nth term of the sequence 4, 7, 10, 13, 16, … Prove that the product of any two terms of this sequence is also a term of the sequence. 4. In this question a and b are numbers where a = b + 2. The sum of a and b is equal to the product a and b. Show that a and b are not integers. 5. Show that (2a – 1)2 – (2b – 1)2 = 4(a – b)(a + b – 1). Prove that the difference between the squares of any two odd numbers is a multiple of 8. 6. Prove that (3n + 1)2 – (3n – 1)2 is a multiple of 4, for all positive integer values of n. Prove that the square of any odd number is always one more than a multiple of 8. 1. The square of an odd number can be written as 2n 1 4n2 4n 1 2. We have to show that this is one more than a multiple of 8. This does not look obvious, although we have the + 1, 4n2 4n 1 , it does not seem as though 4n2 4n is a multiple of 8. 3. But 4n2 4n 4 n2 n . So if n was even then n 2 n would also be even, and so we would an 2 even number multiplied by 4 which is a multiple of 8. 4. If n was odd then we would have to consider that for n 2 n , an odd number squared is odd plus an odd number is even. So again we have an even number multiplied by 4 which is a multiple of 8. 1. Show that the sum of any three consecutive multiples of 3 is also a multiple of 3. 3n 3n 3 3n 6 9n 9 2. But we have to show that 9n 9 is a multiple of 3, i.e. of the form 3 3. So factorise 9n 9 3 3n 3 4. Since 3n 3 is any whole number. Then to multiply it by 3 gives a multiple of 3. 1. 2. Write down the nth term of the sequence 4, 7, 10, 13, 16, … Prove that the product of any two terms of this sequence is also a term of the sequence. nth term of the sequence is: 3n 1 We are asked to show that the product of any two terms of this sequence. Therefore it will not be right to start with 3n 1 3n 1 as these are identical terms of the sequence. So introduce a new variable. 3n 1 3k 1 3. OUR GOAL is to show that the product is a term of the sequence, i.e. of the form 3n 1 or more generally 3 1 4. So manipulate 3n 1 3k 1 5. 3n 13k 1 9nk 3n 3k 1 6. 3 7. 3 3nk n k 1 , as required. 1. In this question a and b are numbers where a = b + 2. The sum of a and b is equal to the product a and b. Show that a and b are not integers. We are told that the sum a and b is equal to the product a and b . That is a b ab Since we know that a b 2 let us substitute this into the equation a b ab so that we have only 2. 1 , we need to take out a factor of 3 and ensure that we have the plus one. one variable. 3. 4. b 2 b b 2 b b 2 b b 2 b 2 This becomes 2b 2 b 2b 2 b2 integer too as a b 2 and we know that square root of 2 is not an integer, so a is not an