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Explain why (n+1)(n+20) is an even number

odd. Explain why (n+1)(n+20) is an even number. even. even. If n is an even number n+1 is n+20 is (n+1)(n+20) is odd x even = If n is an odd number n+1 is n+20 is (n+1)(n+20) is even x odd = n can only be odd or even and BOTH CASES GIVES AN EVEN ANSWER. even. odd. even.

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Explain why (n+1)(n+20) is an even number

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  1. odd Explain why (n+1)(n+20) is an even number even even If n is an even number n+1 is n+20 is (n+1)(n+20) is odd x even = If n is an odd number n+1 is n+20 is (n+1)(n+20) is even x odd = n can only be odd or even and BOTH CASES GIVES AN EVEN ANSWER. even odd even

  2. Prove that the difference between the squares of any two consecutive numbers is odd • Let the consecutive numbers be n and n+1 • (n)² = n² • (n+1)² = (n+1)(n+1) = n²+2n+1 • Difference between the squares = n²+2n+1-n² = 2n+1 2n+1 is odd for all n 

  3. Prove that the difference between the squares of any two consecutive odd numbers is a multiple of 8 • Let the odd numbers be 2n+1 and 2n+3 • (2n+1)²= (2n+1)(2n+1) = 4n²+4n+1 • (2n+3)² = (2n+3)(2n+3) = 4n²+12n+9 • Difference between the squares = 4n²+12n+9-(4n²+4n+1) = 4n²+12n+9-4n²-4n-1 = 8n+8 = 8(n+1) Careful with the negative signs – use brackets!! Always end up with an expression to factorise OOH – I do like those 8’s!!! 

  4. Show that (2a-1)² - (2b-1)² = 4(a-b)(a+b-1) • (2a-1)² = (2a-1)(2a-1) = 4a²-4a+1 • (2b-1)² = (2b-1)(2b-1) = 4b²-4b+1 • (2a-1)² - (2b-1)² = 4a²-4a+1-(4b²-4b+1) = 4a²-4a+1-4b²+4b-1 = 4a²-4a-4b²+4b = 4(a²-b²+b-a) = 4((a-b)(a+b)+(b-a)) = 4((a-b)(a+b)-(a-b)) = 4(a-b)(a+b-1) Careful with the negative signs – use brackets!! OOH! – THERE’S A HIDDEN FACTOR!! 

  5. Prove that the difference between the squares of any two odd numbers is a multiple of 8 • Let the odd numbers be 2r-1 and 2p-1 • We know (2r-1)² - (2p-1)² = 4(r-p)(r+p-1) • If r and p are both even or both odd • r-p is • r+p-1 is • 4(r-p)(r+p-1) = 4(even)(odd) = 4(even) = 4(2n) = 8n = multiple of 8 even odd 

  6. odd If one of r and p is odd • r-p is • r+p-1 is • 4(r-p)(r+p-1) = 4(odd)(even) = 4(even) = 4(2n) = 8n = mulitple of 8 FOR ALL THE POSSIBLE COMBINATIONS OF r AND p, WE ALWAYS GET A MULTIPLE OF 8 even 

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