| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove summation formula |
| Difficulty | Standard +0.3 This is a standard proof by induction for a summation formula with straightforward algebra. The base case is trivial, and the inductive step requires expanding (2(k+1)-1)² and factoring the resulting cubic, which are routine A-level techniques. Slightly easier than average as it follows the standard induction template without requiring creative insight. |
| Spec | 4.01a Mathematical induction: construct proofs4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| If \(n=1\): \(\sum_{r=1}^{1}(2r-1)^2 = (2-1)^2 = 1\) and \(\frac{1}{3}n(4n^2-1) = \frac{1}{3}(1)(4(1)^2-1) = 1\); (LHS=RHS) so true for \(n=1\) | B1 | Both sides clearly evaluated as 1; conclusion "true for \(n=1\)" must be present |
| Assume true for \(n=k\) so \(\sum_{r=1}^{k}(2r-1)^2 = \frac{1}{3}k(4k^2-1)\); then \(\sum_{r=1}^{k+1}(2r-1)^2 = \frac{1}{3}k(4k^2-1) + \left(2(k+1)-1\right)^2\) | M1 | Assumes result for \(n=k\) and attempts to add \((k+1)\)th term |
| E.g. \(= \frac{1}{3}(2k+1)(2k^2+5k+3)\) or \(= \frac{4}{3}k^3 - \frac{1}{3}k + 4k^2 + 4k + 1\) | dM1 | Makes progress by factorising common factor \((2k+1)\) or expanding fully |
| \(\frac{1}{3}(k+1)(2k+3)(2k+1)\) or \(\frac{1}{3}(k+1)(4k^2+8k+3)\) or \(\frac{4k^3}{3} + 4k^2 + \frac{11k}{3} + 1\) | A1 | Correct factorised or expanded form with \((k+1)\) as factor; from correct work |
| \(= \frac{1}{3}(k+1)(4(k+1)^2 - 1)\) | A1 | Completes inductive step correctly; no errors |
| If true for \(n=k\) then shown true for \(n=k+1\); true for \(n=1\); therefore true for all positive integers \(n\) | A1 | Complete conclusion including "true for \(n=1\)", "true for \(n=k\) implies true for \(n=k+1\)", "true for all \(n\)" |
# Question 6:
| Working | Mark | Guidance |
|---------|------|----------|
| If $n=1$: $\sum_{r=1}^{1}(2r-1)^2 = (2-1)^2 = 1$ and $\frac{1}{3}n(4n^2-1) = \frac{1}{3}(1)(4(1)^2-1) = 1$; **(LHS=RHS)** so true for $n=1$ | B1 | Both sides clearly evaluated as 1; conclusion "true for $n=1$" must be present |
| Assume true for $n=k$ so $\sum_{r=1}^{k}(2r-1)^2 = \frac{1}{3}k(4k^2-1)$; then $\sum_{r=1}^{k+1}(2r-1)^2 = \frac{1}{3}k(4k^2-1) + \left(2(k+1)-1\right)^2$ | M1 | Assumes result for $n=k$ and attempts to add $(k+1)$th term |
| E.g. $= \frac{1}{3}(2k+1)(2k^2+5k+3)$ or $= \frac{4}{3}k^3 - \frac{1}{3}k + 4k^2 + 4k + 1$ | dM1 | Makes progress by factorising common factor $(2k+1)$ or expanding fully |
| $\frac{1}{3}(k+1)(2k+3)(2k+1)$ or $\frac{1}{3}(k+1)(4k^2+8k+3)$ or $\frac{4k^3}{3} + 4k^2 + \frac{11k}{3} + 1$ | A1 | Correct factorised or expanded form with $(k+1)$ as factor; from correct work |
| $= \frac{1}{3}(k+1)(4(k+1)^2 - 1)$ | A1 | Completes inductive step correctly; no errors |
| If true for $n=k$ then shown true for $n=k+1$; true for $n=1$; therefore true for all positive integers $n$ | A1 | Complete conclusion including "true for $n=1$", "true for $n=k$ implies true for $n=k+1$", "true for all $n$" |\begin{enumerate}
\item Prove by induction that, for all positive integers $n$,
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } ( 2 r - 1 ) ^ { 2 } = \frac { 1 } { 3 } n \left( 4 n ^ { 2 } - 1 \right)$$
\hfill \mbox{\textit{Edexcel CP1 2024 Q6 [6]}}