FP1 June 2015 Q6
6.
| Scheme | Marks |
|---|---|
| If \(n = 1\), \(\begin{pmatrix} 1 & 0 \\ -1 & 5 \end{pmatrix}^1 = \begin{pmatrix} 1 & 0 \\ -\tfrac{1}{4}(5^{1} - 1) & 5^{1} \end{pmatrix}\) so true for \(\boldsymbol{n = 1}\) | B1 |
| Assume result true for \(n = k\) | |
| \(\begin{pmatrix} 1 & 0 \\ -1 & 5 \end{pmatrix}^{k+1} = \begin{pmatrix} 1 & 0 \\ -\tfrac{1}{4}(5^{k} - 1) & 5^{k} \end{pmatrix}\begin{pmatrix} 1 & 0 \\ -1 & 5 \end{pmatrix}\) or \(\begin{pmatrix} 1 & 0 \\ -1 & 5 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ -\tfrac{1}{4}(5^{k} - 1) & 5^{k} \end{pmatrix}\) | M1 |
| \(\begin{pmatrix} 1 & 0 \\ -1 & 5 \end{pmatrix}^{k+1} = \begin{pmatrix} 1 & 0 \\ -\tfrac{1}{4}(5^k - 1) - 5^k & 5 \times 5^k \end{pmatrix}\) or \(\begin{pmatrix} 1 & 0 \\ -1 - 5.\tfrac{1}{4}(5^k - 1) & 5 \times 5^k \end{pmatrix}\) | M1 A1 |
| \(= \begin{pmatrix} 1 & 0 \\ -\dfrac{1}{4}5^k + \dfrac{1}{4} - 5^k & 5^{k+1} \end{pmatrix}\) or \(\begin{pmatrix} 1 & 0 \\ -1 - \tfrac{1}{4}5^{k+1} + \dfrac{5}{4} & 5^{k+1} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ -\tfrac{1}{4}(5^{k+1} - 1) & 5^{k+1} \end{pmatrix}\) | A1 |
| True for \(\boldsymbol{n = k + 1}\) if true for \(\boldsymbol{n = k}\), (and true for \(\boldsymbol{n = 1}\)) so true by induction for all \(\boldsymbol{n \in \mathbb{Z}^{+}}\). | A1cso |
| (6) |
Notes
B1: Checks \(n = 1\) on both sides and states true for \(n = 1\) seen anywhere.
M1: Assumes true for \(n = k\) and indicates intention to multiply power \(k\) by power 1 either way around.
M1: Multiplies matrices. Condone one slip. A1: Correct unsimplified matrix
A1: Intermediate step required cao
A1: cso Makes correct induction statement including at least statements in bold.
Statement true for \(\boldsymbol{n = 1}\) here could contribute to B1 mark earlier.
| Scheme | Marks |
|---|---|
| If \(n = 1\), \(\displaystyle\sum_{r=1}^{n} (2r - 1)^2 = 1\) and \(\dfrac{1}{3}n(4n^2 - 1) = 1\), so true for \(\boldsymbol{n = 1}\). | B1 |
| Assume result true for \(n = k\) so \(\displaystyle\sum_{r=1}^{k+1} (2r - 1)^2 = \frac{1}{3}k(4k^2 - 1) + (2(k + 1) - 1)^2\) | M1 |
| \(= \displaystyle\sum_{r=1}^{k+1} (2r - 1)^2 = \frac{1}{3}(2k + 1)\{(2k^2 - k) + (3(2k + 1))\}\) | M1 A1 |
| \(= \dfrac{1}{3}(2k + 1)\{(2k^2 + 5k + 3)\}\) or \(\dfrac{1}{3}(k + 1)(4k^2 + 8k + 3)\) or \(\dfrac{1}{3}((2k + 3)(2k^2 + 3k + 1))\) | |
| \(= \dfrac{1}{3}(k + 1)(2k + 1)(2k + 3) \quad = \dfrac{1}{3}(k + 1)(4(k + 1)^2 - 1)\) | dA1 |
| True for \(\boldsymbol{n = k + 1}\) if true for \(\boldsymbol{n = k}\), (and true for \(\boldsymbol{n = 1}\)) so true by induction for all \(\boldsymbol{n \in \mathbb{Z}^{+}}\) | A1cso |
| (6) | |
| (12 marks) |
Notes
B1: Checks \(n = 1\) on both sides and states true for \(n = 1\) seen anywhere.
M1: Assumes true for \(n = k\) and adds \((k+1)^{\text{th}}\) term to sum of \(k\) terms. Accept \(4(k + 1)^2 - 4(k + 1) + 1\) or \((2k + 1)^2\) for \((k+1)^{\text{th}}\) term. M1: Factorises out a linear factor of the three possible - usually \(2k+1\)
A1: Correct expression with one linear and one quadratic factor.
dA1: Need to see \(\dfrac{1}{3}(k + 1)(4(k + 1)^2 - 1)\) somewhere dependent upon previous A1.
Accept assumption plus \((k+1)^{\text{th}}\) term and \(\dfrac{1}{3}(k + 1)(4(k + 1)^2 - 1)\) both leading to \(\dfrac{1}{3}(4k^3 + 12k^2 + 11k + 3)\) then award for expressions seen as above.
A1: cso Makes correct complete induction statement including at least statements in bold. Statement true for \(n = 1\) here could contribute to B1 mark earlier.