Moderate -0.5 This is a straightforward algebraic proof requiring factorization of n³-n = n(n-1)(n+1) and recognizing that consecutive integers guarantee an even product. While it requires proof technique rather than just calculation, it's a standard C1-level proof that's simpler than typical multi-step problems, making it slightly easier than average.
Question 8:
8 | any general attempt at n being odd
and n being even even
n odd implies n3 odd and odd − odd
= even
n even implies n3 even and even −
even = even | M1
A1
A1 | M0 for just trying numbers, even if some
odd, some even
or n(n2 − 1) used with n odd implies n2 −
1 even and odd × even = even etc
[allow even × odd = even]
or A2 for n(n − 1)(n + 1) = product of 3
consecutive integers; at least one even
so product even;
odd3 − odd = odd etc is not sufft for A1
SC1 for complete general method for
only one of odd or even eg n = 2m
leading to 2(4m3 − m) | 3