4MA0 Higher June 2017 Paper 3R Q20
20 \(k = 2^p - 1\) where \(p\) is an integer \(\gt 1\)
\(N = k^2 - 1\)
Show that \(2^{p+1}\) is a factor of \(N\)
(3)
| Scheme | Marks |
|---|---|
| \((2^p)^2 - 2^p - 2^p + 1\) or \(2^{2p} - 2^p - 2^p + 1\) or \(4^p - 2^p - 2^p + 1\) | M2 |
| \(N = 2^{p+1}(2^{p-1} - 1)\) Answer: Shown | A1 |
| (3) | |
| (3 marks) |
Notes
M2: for correct expansion
If not M2 then award M1 for
3 terms correct from \((2^p)^2 - 2^p - 2^p + 1\)
A1: for correct factorised expression from correct working (dep on M2)
Alternative method
| Scheme | Marks |
|---|---|
| \(N = (k - 1)(k + 1)\) | M1 |
| \(N = (2^p - 1 - 1)(2^p - 1 + 1)\) | M1 |
| \(N = 2^{p+1}(2^{p-1} - 1)\) Answer: Shown | A1 |
Notes
M1: for correct factorisation
M1: for correct factorisation and substitution (implies B1)
A1: for correct factorised expression (dep on M2)