Foundation November 2022 Paper 1 Q19
19 \(n\) is an odd number.
Why is \(\quad n(n + 1) \quad\) always an even number? [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(n + 1\) is even and odd \(\times\) even is even | B2 | oe B1 \(n + 1\) is even or odd \(\times\) even is even |
| Alternative method 2 | ||
| \(n^2 + n\) and odd\(^2\) is odd and odd + odd is even | B2 | oe B1 \(n^2 + n\) or odd\(^2\) is odd and odd + odd is even |
| Alternative method 3 | ||
| \(n\) and \(n + 1\) are consecutive numbers and odd \(\times\) even is even | B2 | oe B1 \(n\) and \(n + 1\) are consecutive numbers or odd \(\times\) even is even |
Additional guidance
| Alt 1 odd + 1 = even and multiplying an odd and an even = even | B2 |