AS June 2018 Paper 1 Q8
8 Prove by induction that \(\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}^n = \begin{pmatrix} 1 & 2^n - 1 \\ 0 & 2^n \end{pmatrix}\) for all positive integers \(n\). [6]
| Scheme | Marks | AO |
|---|---|---|
| When \(n = 1\), \(\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}^1 = \begin{pmatrix} 1 & 2^1 - 1 \\ 0 & 2^1 \end{pmatrix}\) | B1 | 2.1 |
| [Assume] \(n = k\): \(\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}^k = \begin{pmatrix} 1 & 2^k - 1 \\ 0 & 2^k \end{pmatrix}\) | M1 | 2.2a |
| Then \(\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}^{k+1} = \begin{pmatrix} 1 & 2^k - 1 \\ 0 & 2^k \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}\) | M1 | 1.1 |
| \(= \begin{pmatrix} 1 & 1 + 2^{k+1} - 2 \\ 0 & 2^{k+1} \end{pmatrix} = \begin{pmatrix} 1 & 2^{k+1} - 1 \\ 0 & 2^{k+1} \end{pmatrix}\) | A1* | 1.1 |
| so if true for \(n = k\) then true for \(n = k + 1\) [as true for \(n = 1\)] therefore true for all \(n\). | M1dep* A1dep* | 2.2a 2.4 |
| [6] |
Notes
M1: (2nd) or \(\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 1 & 2^k - 1 \\ 0 & 2^k \end{pmatrix}\)
A1*: or \(\begin{pmatrix} 1 & 2^k - 1 + 2^k \\ 0 & 2^{k+1} \end{pmatrix}\)
M1dep*: must clearly ‘assume’ \(n = k\)
‘true for \(n = k\) and \(k + 1\)’ is M0
A1dep*: must have established truth for \(n = 1\) for final mark
but need not re-state it at the end