FP1 June 2017 Q9
9.
| Scheme | Marks |
|---|---|
| \(u_{n+2} = 6u_{n+1} - 9u_n,\ n \geqslant 1,\ u_1 = 6,\ u_2 = 27;\quad u_n = 3^n(n + 1)\) | |
| \(n = 1;\quad u_1 = 3(2) = 6\) \(n = 2;\quad u_2 = 3^2(2 + 1) = 27\) So \(u_n\) is true when \(n = 1\) and \(n = 2\). | B1 |
| Assume that \(u_k = 3^k(k + 1)\) and \(u_{k+1} = 3^{k+1}(k + 2)\) are true. | |
| Then \(u_{k+2} = 6u_{k+1} - 9u_k\) \(= 6(3^{k+1})(k + 2) - 9(3^k)(k + 1)\) | M1 A1 |
| \(= 2(3^{k+2})(k + 2) - (3^{k+2})(k + 1)\) | M1 |
| \(= (3^{k+2})(2k + 4 - k - 1)\) \(= (3^{k+2})(k + 3)\) | |
| \(= (3^{k+2})(k + 2 + 1)\) | A1 |
| If the result is true for \(n = k\) and \(n = k+1\) then it is now true for \(n = k+2\). As it is true for \(n = 1\) and \(n = 2\) then it is true for all \(n\) \((\in \mathbb{Z}^+)\). | A1 cso |
| (6) |
Notes
B1: Check that \(u_1 = 6\) and \(u_2 = 27\)
Could assume for \(n = k, n = k - 1\) and show for \(n = k + 1\)
M1: Substituting \(u_k\) and \(u_{k+1}\) into \(u_{k+2} = 6u_{k+1} - 9u_k\)
A1: Correct expression
M1: Achieves an expression in \(3^{k+2}\)
A1: \((3^{k+2})(k + 2 + 1)\) or \((3^{k+2})(k + 3)\)
A1 cso: Correct conclusion seen at the end. Condone true for \(n = 1\) and \(n = 2\) seen anywhere. This should be compatible with assumptions.
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(n) = 3^{3n-2} + 2^{3n+1}\) is divisible by 19 | |
| Way 1 | |
| \(\mathrm{f}(1) = 3^1 + 2^4 = 19\) {which is divisible by 19}. \(\{\therefore \mathrm{f}(n)\) is divisible by 19 when \(n = 1\}\) | B1 |
| {Assume that for \(n = k\), \(\mathrm{f}(k) = 3^{3k-2} + 2^{3k+1}\) is divisible by 19 for \(k \in \mathbb{Z}^+\).} | |
| \(\mathrm{f}(k + 1) - \mathrm{f}(k) = 3^{3(k+1)-2} + 2^{3(k+1)+1} - (3^{3k-2} + 2^{3k+1})\) | M1 |
| \(\mathrm{f}(k + 1) - \mathrm{f}(k) = 27(3^{3k-2}) + 8(2^{3k+1}) - (3^{3k-2} + 2^{3k+1})\) \(\mathrm{f}(k + 1) - \mathrm{f}(k) = 26(3^{3k-2}) + 7(2^{3k+1})\) \(= 7(3^{3k-2} + 2^{3k+1}) + 19(3^{3k-2})\) or \(= 26(3^{3k-2} + 2^{3k+1}) - 19(2^{3k+1})\) \(= 7\mathrm{f}(k) + 19(3^{3k-2})\) or \(= 26\mathrm{f}(k) - 19(2^{3k+1})\) | A1; A1 |
| \(\therefore \mathrm{f}(k + 1) = 8\mathrm{f}(k) + 19(3^{3k-2})\) or \(\mathrm{f}(k + 1) = 27\mathrm{f}(k) - 19(2^{3k+1})\) | dM1 |
| \(\{\therefore \mathrm{f}(k + 1) = 8\mathrm{f}(k) + 19(3^{3k-2})\) is divisible by 19 as both \(8\mathrm{f}(k)\) and \(19(3^{3k-2})\) are both divisible by 19} | |
| If the result is true for \(n = k\), then it is now true for \(n = k + 1\). As the result has shown to be true for \(n = 1\), then the result is true for all \(n\) \((\in \mathbb{Z}^+)\). | A1 cso |
| (6) | |
| (12 marks) |
Notes
In all ways, first M is for applying \(\mathrm{f}(k + 1)\) with at least 1 power correct. The second M is dependent on at least one accuracy being awarded and making \(\mathrm{f}(k + 1)\) the subject and the final A is correct solution only.
B1: Shows \(\mathrm{f}(1) = 19\)
M1: Applies \(\mathrm{f}(k + 1)\) with at least 1 power correct
A1; A1: Either \(7(3^{3k-2} + 2^{3k+1})\) or \(7\mathrm{f}(k);\ 19(3^{3k-2})\)
or \(26(3^{3k-2} + 2^{3k+1})\) or \(26\mathrm{f}(k);\ -19(2^{3k+1})\)
dM1: Dependent on at least one of the previous accuracy marks being awarded. Makes Applies \(\mathrm{f}(k + 1)\) with at least 1 power correct the subject
A1 cso: Correct conclusion seen at the end. Condone true for \(n = 1\) stated earlier.
Note: Accept use of \(\mathrm{f}(k) = 3^{3k-2} + 2^{3k+1} = 19m\) o.e. and award method and accuracy as above.
(ii) Way 2
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(1) = 3^1 + 2^4 = 19\) {which is divisible by 19}. \(\{\therefore \mathrm{f}(n)\) is divisible by 19 when \(n = 1\}\) | B1 |
| Assume that for \(n = k\), \(\mathrm{f}(k) = 3^{3k-2} + 2^{3k+1}\) is divisible by 19 for \(k \in \mathbb{Z}^+\). | |
| \(\mathrm{f}(k + 1) = 3^{3(k+1)-2} + 2^{3(k+1)+1}\) | M1 |
| \(\mathrm{f}(k + 1) = 27(3^{3k-2}) + 8(2^{3k+1})\) \(= 8(3^{3k-2} + 2^{3k+1}) + 19(3^{3k-2})\) or \(= 27(3^{3k-2} + 2^{3k+1}) - 19(2^{3k+1})\) | A1; A1 |
| \(\therefore \mathrm{f}(k + 1) = 8\mathrm{f}(k) + 19(3^{3k-2})\) or \(\mathrm{f}(k + 1) = 27\mathrm{f}(k) - 19(2^{3k+1})\) | dM1 |
| \(\{\therefore \mathrm{f}(k + 1) = 8\mathrm{f}(k) + 19(3^{3k-2})\) is divisible by 19 as both \(8\mathrm{f}(k)\) and \(19(3^{3k-2})\) are both divisible by 19} | |
| If the result is true for \(n = k\), then it is now true for \(n = k + 1\). As the result has shown to be true for \(n = 1\), then the result is true for all \(n\) \((\in \mathbb{Z}^+)\). | A1 cso |
| (6) |
B1: Shows \(\mathrm{f}(1) = 19\)
M1: Applies \(\mathrm{f}(k + 1)\) with at least 1 power correct
A1; A1: Either \(8(3^{3k-2} + 2^{3k+1})\) or \(8\mathrm{f}(k);\ 19(3^{3k-2})\)
or \(27(3^{3k-2} + 2^{3k+1})\) or \(27\mathrm{f}(k);\ -19(2^{3k+1})\)
dM1: Dependent on at least one of the previous accuracy marks being awarded.
A1 cso: Correct conclusion seen at the end. Condone true for \(n = 1\) stated earlier.
(ii) Way 3
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(1) = 3^1 + 2^4 = 19\) {which is divisible by 19}. \(\{\therefore \mathrm{f}(n)\) is divisible by 19 when \(n = 1\}\) | B1 |
| Assume that for \(n = k\), \(\mathrm{f}(k) = 3^{3k-2} + 2^{3k+1}\) is divisible by 19 for \(k \in \mathbb{Z}^+\). | |
| \(\mathrm{f}(k + 1) - \alpha\mathrm{f}(k) = 3^{3(k+1)-2} + 2^{3(k+1)+1} - \alpha(3^{3k-2} + 2^{3k+1})\) | M1 |
| \(\mathrm{f}(k + 1) - \alpha\mathrm{f}(k) = (27 - \alpha)(3^{3k-2}) + (8 - \alpha)2^{3k+1}\) \(= (8 - \alpha)(3^{3k-2} + 2^{3k+1}) + 19(3^{3k-2})\) or \(= (27 - \alpha)(3^{3k-2} + 2^{3k+1}) - 19(2^{3k+1})\) | A1; A1 |
| \(\therefore \mathrm{f}(k + 1) = 8\mathrm{f}(k) + 19(3^{3k-2})\) or \(\mathrm{f}(k + 1) = 27\mathrm{f}(k) - 19(2^{3k+1})\) | dM1 |
| \(\{\therefore \mathrm{f}(k + 1) = 27\mathrm{f}(k) - 19(2^{3k+1})\) is divisible by 19 as both \(27\mathrm{f}(k)\) and \(19(2^{3k+1})\) are both divisible by 19} | |
| If the result is true for \(n = k\), then it is now true for \(n = k + 1\). As the result has shown to be true for \(n = 1\), then the result is true for all \(n\) \((\in \mathbb{Z}^+)\). | A1 cso |
| (6) |
B1: Shows \(\mathrm{f}(1) = 19\)
M1: Applies \(\mathrm{f}(k + 1)\) with at least 1 power correct
A1; A1: \((8 - \alpha)(3^{3k-2} + 2^{3k+1})\) or \((8 - \alpha)\mathrm{f}(k);\ 19(3^{3k-2})\). NB choosing \(\alpha = 8\) makes first term disappear.
\((27 - \alpha)(3^{3k-2} + 2^{3k+1})\) or \((27 - \alpha)\mathrm{f}(k);\ -19(2^{3k+1})\). NB choosing \(\alpha = 27\) makes first term disappear.
dM1: Dependent on at least one of the previous accuracy marks being awarded. Makes \(\mathrm{f}(k + 1)\) the subject.
A1 cso: Correct conclusion seen at the end. Condone true for \(n = 1\) stated earlier.