A6Algebraic proof

• C1 for \(2n\) is 2 × an integer / a multiple of 2
• C1 for \(2n + 1\) is 1 more than an even number

Worked solution: Any integer multiplied by 2 is a multiple of 2, so \(2n\) is even. Adding 1 to an even number always gives an odd number, so \(2n + 1\) is odd.

• B1 for \(10a + b\) or \(10b + a\)
• M1 for \((10a + b) - (10b + a)\)
• C1 for \(9(a - b)\) with a conclusion

Worked solution: The numbers are \(10a + b\) and \(10b + a\).
\((10a + b) - (10b + a) = 9a - 9b = 9(a - b)\), which is a multiple of 9.

• M1 for two consecutive even numbers, e.g. \(2n\) and \(2n + 2\)
• M1 for \(4n^2 + 4n\) or \(4n(n + 1)\)
• C1 for stating that one of \(n\) and \(n + 1\) must be even, so \(n(n + 1)\) is even
• C1 for concluding \(4 \times \text{even} = 8k\), a multiple of 8

Worked solution: \(2n(2n + 2) = 4n^2 + 4n = 4n(n + 1)\).
\(n\) and \(n + 1\) are consecutive integers, so one of them is even and \(n(n + 1) = 2k\) for some integer \(k\).
So the product is \(4 \times 2k = 8k\), a multiple of 8.