AS June 2024 Paper 1 Q7
7.
| Scheme | Marks | AO |
|---|---|---|
| \(n = 1,\quad lhs = \dfrac{1}{1(2)} = \dfrac{1}{2}\quad rhs = \dfrac{1}{1 + 1} = \dfrac{1}{2}\) So the result is true for \(n = 1\) | B1 | 2.2a |
| Assume true for \(n = k\) so \(\displaystyle\sum_{r=1}^{k} \frac{1}{r(r + 1)} = \frac{k}{k + 1}\) | M1 | 2.4 |
| \(\displaystyle\sum_{r=1}^{k+1} \frac{1}{r(r + 1)} = \frac{k}{k + 1} + \frac{1}{(k + 1)(k + 2)} = \frac{k(k + 2)}{(k + 1)(k + 2)} + \frac{1}{(k + 1)(k + 2)}\) | M1 | 2.1 |
| \(= \dfrac{k(k + 2) + 1}{(k + 1)(k + 2)} = \dfrac{k^2 + 2k + 1}{(k + 1)(k + 2)} = \dfrac{(k + 1)^2}{(k + 1)(k + 2)} = \dfrac{\underline{k + 1}}{\underline{k + 1} + 1}\) cso | A1 | 1.1b |
| If true for \(\boldsymbol{n = k}\) then it has been shown true for \(\boldsymbol{n = k + 1}\) and as it is true for \(\boldsymbol{n = 1}\), the statement is true for all positive integers \(\boldsymbol{n}\). | A1 | 2.4 |
| (5) |
Notes
(i) Must be using induction to score any marks
B1: Shows that the result holds for \(n = 1\). Minimum LHS = \(\dfrac{1}{1(2)}\) RHS \(= \dfrac{1}{1 + 1}\)
M1: Makes a statement that assumes the result is true for some value of \(n\), say \(k\)
M1: Attempts to add the next term and makes progress by attempting a common denominator.
A1: Achieves a correct expression in terms of \(k + 1\), with no errors seen cso. Alternatively finds \(\displaystyle\sum_{r=1}^{k+1} \frac{1}{r(r + 1)} = \frac{k + 1}{k + 2}\) and reaches the same expression using induction.
A1: Correct conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all four bold points either at the end of their solution or as a narrative in their solution. Not dependent on B1 as long as attempted
Way 1: f(\(\boldsymbol{k}\) + 1)
| Scheme | Marks | AO |
|---|---|---|
| \(n = 1,\quad 3^{2n+4} - 2^{2n} = 3^6 - 2^2 = 729 - 4 = 725\) So the result is true for \(n = 1\) as 725 is divisible by 5 | B1 | 2.2a |
| Assume true for \(n = k\) so is \(3^{2k+4} - 2^{2k}\) divisible by 5 | M1 | 2.4 |
| \(\mathrm{f}(k + 1) = 3^{2(k+1)+4} - 2^{2(k+1)} = 3^{2k+6} - 2^{2k+2}\) Look for \(A \times 3^{2k+4} - A \times 2^{2k} + B \times 2^{2k}\) or \(A \times 3^{2k+4} - A \times 2^{2k} + B \times 3^{2k+4}\) \(9 \times 3^{2k+4} - 9 \times 2^{2k} + 5 \times 2^{2k}\) or \(4 \times 3^{2k+4} - 4 \times 2^{2k} + 5 \times 3^{2k+4}\) | M1 | 2.1 |
| \(= 9\mathrm{f}(k) + 5 \times 2^{2k}\) or \(= 4\mathrm{f}(k) + 5 \times 3^{2k+4}\) | A1 | 1.1b |
| If true for \(\boldsymbol{n = k}\) then it has been shown true for \(\boldsymbol{n = k + 1}\) and as it is true for \(\boldsymbol{n = 1}\), the statement is true for all positive integers \(\boldsymbol{n}\). | A1 | 2.4 |
| (5) | ||
| (10 marks) |
Notes
Way 1: f(\(\boldsymbol{k}\) + 1)
B1: Shows that the result holds for \(n = 1\)
M1: Makes a statement that assumes the result is true for some value of \(n\), say \(k\)
M1: Attempts f (\(k\) + 1) and attempts to express in terms of f (\(k\))
A1: Achieves a correct expression in terms of f (\(k\))
A1: Correct conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all four bold points either at the end of their solution or as a narrative in their solution. Not dependent on B1 as long as attempted
Alternative: Way 2: f(\(\boldsymbol{k}\) + 1) − f(\(\boldsymbol{k}\))
| Scheme | Marks | AO |
|---|---|---|
| \(n = 1,\quad 3^{2n+4} - 2^{2n} = 3^6 - 2^2 = 729 - 4 = 725\) So the result is true for \(n = 1\) as 725 is divisible by 5 | B1 | 2.2a |
| Assume true for \(n = k\) so \(3^{2k+4} - 2^{2k}\) is divisible by 5 | M1 | 2.4 |
| \(\mathrm{f}(k + 1) - \mathrm{f}(k) = 3^{2(k+1)+4} - 2^{2(k+1)} - 3^{2k+4} + 2^{2k}\) Look for \(A \times 3^{2k+4} - A \times 2^{2k} + B \times 2^{2k}\) or \(A \times 3^{2k+4} - A \times 2^{2k} + B \times 3^{2k+4}\) \(= 8 \times 3^{2k+4} - 8 \times 2^{2k} + 5 \times 2^{2k}\) or \(3 \times 3^{2k+4} - 3 \times 2^{2k} + 5 \times 3^{2k+4}\) | M1 | 2.1 |
| \(\mathrm{f}(k + 1) = 9\mathrm{f}(k) + 5 \times 2^{2k}\) or \(\mathrm{f}(k + 1) = 4\mathrm{f}(k) + 5 \times 3^{2k+4}\) | A1 | 1.1b |
| If true for \(\boldsymbol{n = k}\) then it has been shown true for \(\boldsymbol{n = k + 1}\) and as it is true for \(\boldsymbol{n = 1}\), the statement is true for all positive integers \(\boldsymbol{n}\). | A1 | 2.4 |
| (5) |
B1: Shows that the result holds for \(n = 1\)
M1: Makes a statement that assumes the result is true for some value of \(n\), say \(k\)
M1: Attempts f (\(k\) + 1) – f (\(k\)) and attempts to express in terms of f (\(k\))
A1: Achieves a correct expression for f (\(k\) + 1) in terms of f (\(k\))
A1: Correct conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all four bold points either at the end of their solution or as a narrative in their solution. Not dependent on B1 as long as attempted
Alternative: Way 3: f(\(\boldsymbol{k}\)) = 5\(\boldsymbol{M}\)
| Scheme | Marks | AO |
|---|---|---|
| \(n = 1,\quad 3^{2n+4} - 2^{2n} = 3^6 - 2^2 = 729 - 4 = 725\) So the result is true for \(n = 1\) as 725 is divisible by 5 | B1 | 2.2a |
| Assume true for \(n = k\) so \(3^{2k+4} - 2^{2k} = 5M\) is divisible by 5 | M1 | 2.4 |
| \(\mathrm{f}(k + 1) = 3^{2(k+1)+4} - 2^{2(k+1)}\ \left(= 3^{2k+6} - 2^{2k+2}\right)\) \(\mathrm{f}(k + 1) = 3^2 \times 3^{2k+4} - 2^2 \times 2^{2k} = 3^2 \times \left(5M + 2^{2k}\right) - 2^2 \times 2^{2k}\) OR \(\mathrm{f}(k + 1) = 3^2 \times 3^{2k+4} - 2^2 \times 2^{2k} = 3^2 \times 3^{2k+4} - 2^2 \times \left(3^{2k+4} - 5M\right)\) | M1 | 2.1 |
| \(\mathrm{f}(k + 1) = 45M + 5 \times 2^{2k}\) OR \(\mathrm{f}(k + 1) = 5 \times 3^{2k+4} + 20M\) o.e. | A1 | 1.1b |
| If true for \(\boldsymbol{n = k}\) then it has been shown true for \(\boldsymbol{n = k + 1}\) and as it is true for \(\boldsymbol{n = 1}\), the statement is true for all positive integers \(\boldsymbol{n}\). | A1 | 2.4 |
| (5) |
B1: Shows that the result holds for \(n = 1\)
M1: Makes a statement that assumes the result is true for some value of \(n\), say \(k\)
M1: Attempts f(\(k\) +1) and writes in terms of 5\(M\).
A1: Achieves a correct expression for f (\(k\) + 1) in terms of \(M\) and \(2^{2k}\) or \(M\) and \(3^{2k+4}\)
A1: Correct conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all four bold points either at the end of their solution or as a narrative in their solution. Not dependent on B1 as long as attempted
Alternative: Way 4: f(\(\boldsymbol{k}\) + 1) − \(\boldsymbol{m}\)f(\(\boldsymbol{k}\))
| Scheme | Marks | AO |
|---|---|---|
| \(n = 1,\quad 3^{2n+4} - 2^{2n} = 3^6 - 2^2 = 729 - 4 = 725\) So the result is true for \(n = 1\) as 725 is divisible by 5 | B1 | 2.2a |
| Assume true for \(n = k\) so \(3^{2k+4} - 2^{2k}\) is divisible by 5 | M1 | 2.4 |
| \(\mathrm{f}(k + 1) - m\mathrm{f}(k) = 3^{2k+6} - 2^{2k+2} - m\left(3^{2k+4} - 2^{2k}\right)\) \(3^2 \times 3^{2k+4} - 2^2 \times 2^{2k} - m \times 3^{2k+4} + m \times 2^{2k}\) \((9 - m) \times 3^{2k+4} - 4 \times 2^{2k} + m \times 2^{2k}\) \((9 - m) \times \left(3^{2k+4} - 2^{2k}\right) + 5 \times 2^{2k}\) | M1 | 2.1 |
| \(\mathrm{f}(k + 1) = (9 - m) \times \mathrm{f}(k) + 5 \times 2^{2k} + m\mathrm{f}(k)\) \(\mathrm{f}(k + 1) = (9 - m) \times \left(3^{2k+4} - 2^{2k}\right) + 5 \times 2^{2k} + m\mathrm{f}(k)\) Note if \(m = 4\) leads to \(5 \times \left(3^{2k+4}\right)\) | A1 | 1.1b |
| If true for \(\boldsymbol{n = k}\) then it has been shown true for \(\boldsymbol{n = k + 1}\) and as it is true for \(\boldsymbol{n = 1}\), the statement is true for all positive integers \(\boldsymbol{n}\). | A1 | 2.4 |
| (5) |
B1: Shows that the result holds for \(n = 1\)
M1: Makes a statement that assumes the result is true for some value of \(n\), say \(k\)
M1: Attempts f (\(k\) + 1) – m f (\(k\)) and attempts to express in terms of f (\(k\))
A1: Achieves a correct expression for f (\(k\) + 1) in terms of f (\(k\))
A1: Correct conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all four bold points either at the end of their solution or as a narrative in their solution. Not dependent on B1 as long as attempted
Note conclusion may be in terms of \(\mathrm{f}(1), \mathrm{f}(k), \mathrm{f}(k + 1)\)