| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove recurrence relation formula |
| Difficulty | Standard +0.3 This is a standard FP1 proof by induction question with two routine parts: (i) proving a recurrence relation formula using the recurrence itself in the inductive step, and (ii) proving divisibility using standard algebraic manipulation. Both follow textbook templates with no novel insights required, making this slightly easier than average A-level difficulty. |
| Spec | 4.01a Mathematical induction: construct proofs4.01b Complex proofs: conjecture and demanding proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(n=1\): \(u_1 = 3(2) = 6\) and \(n=2\): \(u_2 = 3^2(2+1) = 27\) | B1 | Check that \(u_1 = 6\) and \(u_2 = 27\) |
| Assume \(u_k = 3^k(k+1)\) and \(u_{k+1} = 3^{k+1}(k+2)\) are true | Could assume for \(n=k, n=k-1\) and show for \(n=k+1\) | |
| \(u_{k+2} = 6u_{k+1} - 9u_k = 6(3^{k+1})(k+2) - 9(3^k)(k+1)\) | M1 | Substituting \(u_k\) and \(u_{k+1}\) into \(u_{k+2} = 6u_{k+1} - 9u_k\); correct expression |
| \(= 2(3^{k+2})(k+2) - (3^{k+2})(k+1)\) | A1, M1 | Achieves an expression in \(3^{k+2}\) |
| \(= (3^{k+2})(2k+4-k-1)\) | ||
| \(= (3^{k+2})(k+3)\) | ||
| \(= (3^{k+2})(k+2+1)\) | A1 | \((3^{k+2})(k+2+1)\) or \((3^{k+2})(k+3)\) |
| Correct conclusion; true for \(n=k\) and \(n=k+1\) implies true for \(n=k+2\); true for \(n=1\) and \(n=2\) therefore true for all \(n \in \mathbb{Z}^+\) | A1, cso | Condone true for \(n=1\) and \(n=2\) seen anywhere. Must be compatible with assumptions. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(1) = 3^1 + 2^4 = 19\) (divisible by 19) | B1 | Shows \(f(1) = 19\) |
| Assume \(f(k) = 3^{3k-2} + 2^{3k+1}\) is divisible by 19 for \(k \in \mathbb{Z}^+\) | ||
| \(f(k+1) - f(k) = 3^{3(k+1)-2} + 2^{3(k+1)+1} - (3^{3k-2} + 2^{3k+1})\) | M1 | Applies \(f(k+1)\) with at least 1 power correct |
| \(f(k+1) - f(k) = 26(3^{3k-2}) + 7(2^{3k+1})\) | ||
| \(= 7(3^{3k-2} + 2^{3k+1}) + 19(3^{3k-2})\) | A1; A1 | Either \(7(3^{3k-2} + 2^{3k+1})\) or \(7f(k)\); \(19(3^{3k-2})\) or \(26(3^{3k-2} + 2^{3k+1})\) or \(26f(k)\); \(-19(2^{3k+1})\) |
| \(\therefore f(k+1) = 8f(k) + 19(3^{3k-2})\) or \(f(k+1) = 27f(k) - 19(2^{3k+1})\) | dM1 | Dependent on at least one previous accuracy mark; makes \(f(k+1)\) the subject |
| \(f(k+1) = 8f(k) + 19(3^{3k-2})\) is divisible by 19 as both \(8f(k)\) and \(19(3^{3k-2})\) are divisible by 19 | ||
| Correct conclusion; true for \(n=k\) implies true for \(n=k+1\); true for \(n=1\) therefore true for all \(n \in \mathbb{Z}^+\) | A1, cso | Condone true for \(n=1\) stated earlier |
# Question 9:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $n=1$: $u_1 = 3(2) = 6$ and $n=2$: $u_2 = 3^2(2+1) = 27$ | B1 | Check that $u_1 = 6$ and $u_2 = 27$ |
| Assume $u_k = 3^k(k+1)$ and $u_{k+1} = 3^{k+1}(k+2)$ are true | | Could assume for $n=k, n=k-1$ and show for $n=k+1$ |
| $u_{k+2} = 6u_{k+1} - 9u_k = 6(3^{k+1})(k+2) - 9(3^k)(k+1)$ | M1 | Substituting $u_k$ and $u_{k+1}$ into $u_{k+2} = 6u_{k+1} - 9u_k$; correct expression |
| $= 2(3^{k+2})(k+2) - (3^{k+2})(k+1)$ | A1, M1 | Achieves an expression in $3^{k+2}$ |
| $= (3^{k+2})(2k+4-k-1)$ | | |
| $= (3^{k+2})(k+3)$ | | |
| $= (3^{k+2})(k+2+1)$ | A1 | $(3^{k+2})(k+2+1)$ or $(3^{k+2})(k+3)$ |
| Correct conclusion; true for $n=k$ and $n=k+1$ implies true for $n=k+2$; true for $n=1$ and $n=2$ therefore true for all $n \in \mathbb{Z}^+$ | A1, cso | Condone true for $n=1$ and $n=2$ seen anywhere. Must be compatible with assumptions. |
**[6 marks]**
## Part (ii):
### Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(1) = 3^1 + 2^4 = 19$ (divisible by 19) | B1 | Shows $f(1) = 19$ |
| Assume $f(k) = 3^{3k-2} + 2^{3k+1}$ is divisible by 19 for $k \in \mathbb{Z}^+$ | | |
| $f(k+1) - f(k) = 3^{3(k+1)-2} + 2^{3(k+1)+1} - (3^{3k-2} + 2^{3k+1})$ | M1 | Applies $f(k+1)$ with at least 1 power correct |
| $f(k+1) - f(k) = 26(3^{3k-2}) + 7(2^{3k+1})$ | | |
| $= 7(3^{3k-2} + 2^{3k+1}) + 19(3^{3k-2})$ | A1; A1 | Either $7(3^{3k-2} + 2^{3k+1})$ or $7f(k)$; $19(3^{3k-2})$ **or** $26(3^{3k-2} + 2^{3k+1})$ or $26f(k)$; $-19(2^{3k+1})$ |
| $\therefore f(k+1) = 8f(k) + 19(3^{3k-2})$ **or** $f(k+1) = 27f(k) - 19(2^{3k+1})$ | dM1 | Dependent on at least one previous accuracy mark; makes $f(k+1)$ the subject |
| $f(k+1) = 8f(k) + 19(3^{3k-2})$ is divisible by 19 as both $8f(k)$ and $19(3^{3k-2})$ are divisible by 19 | | |
| Correct conclusion; true for $n=k$ implies true for $n=k+1$; true for $n=1$ therefore true for all $n \in \mathbb{Z}^+$ | A1, cso | Condone true for $n=1$ stated earlier |
**[6 marks]**
### Ways 2 & 3 follow the same mark allocation (B1, M1, A1, A1, dM1, A1, cso) with equivalent working.**
**Note:** Accept use of $f(k) = 3^{3k-2} + 2^{3k+1} = 19m$ and award method and accuracy as above.
**[Total: 12 marks]**9. (i) A sequence of numbers is defined by
$$\begin{gathered}
u _ { 1 } = 6 , \quad u _ { 2 } = 27 \\
u _ { n + 2 } = 6 u _ { n + 1 } - 9 u _ { n } \quad n \geqslant 1
\end{gathered}$$
Prove by induction that, for $n \in \mathbb { Z } ^ { + }$
$$u _ { n } = 3 ^ { n } ( n + 1 )$$
(ii) Prove by induction that, for $n \in \mathbb { Z } ^ { + }$
$$f ( n ) = 3 ^ { 3 n - 2 } + 2 ^ { 3 n + 1 } \text { is divisible by } 19$$
\includegraphics[max width=\textwidth, alt={}, center]{536d7ec7-91b0-4fda-a485-2ac4a72c7d59-29_56_20_109_1950}\\
\begin{center}
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\hfill \mbox{\textit{Edexcel FP1 2017 Q9 [12]}}