A2 June 2024 Paper 1 Q6
6. Prove by induction that, for all positive integers \(n\),
\[\sum_{r=1}^{n} (2r - 1)^2 = \frac{1}{3}n\left(4n^2 - 1\right)\](6)
| Scheme | Marks | AO |
|---|---|---|
| If \(n = 1\) \(\displaystyle\sum_{r=1}^{n} (2r - 1)^2 = (2 - 1)^2 = 1\) and \(\dfrac{1}{3}n\left(4n^2 - 1\right) = \dfrac{1}{3}(1)\left(4(1)^2 - 1\right) = 1\) (LHS=RHS) so true for \(n = 1\) | B1 | 2.4 |
| (Assume true for \(n = k\) so \(\displaystyle\sum_{r=1}^{k} (2r - 1)^2 = \dfrac{1}{3}k\left(4k^2 - 1\right)\) then) \(\displaystyle\sum_{r=1}^{k+1} (2r - 1)^2 = \dfrac{1}{3}k\left(4k^2 - 1\right) + \bigl(2(k + 1) - 1\bigr)^2\) | M1 | 2.1 |
| E.g \(= \dfrac{1}{3}(2k + 1)\left(2k^2 + 5k + 3\right)\) or \(= \dfrac{4}{3}k^3 - \dfrac{1}{3}k + 4k^2 + 4k + 1\) | dM1 | 1.1b |
| \(\dfrac{1}{3}(k + 1)(2k + 3)(2k + 1)\) or \(\dfrac{1}{3}(k + 1)\left(4k^2 + 8k + 3\right)\) or \(\dfrac{4k^3}{3} + 4k^2 + \dfrac{11k}{3} + 1\) | A1 | 1.1b |
| \(= \dfrac{1}{3}(k + 1)\left(4(k + 1)^2 - 1\right)\) Or see notes | A1 | 2.2a |
| If the statement is true for \(n = k\) then it has been shown true for \(n = k + 1\) and as it is true for \(n = 1\), the statement is true for all positive integers \(n\). | A1 | 2.4 |
| (6) | ||
| (6 marks) |
Notes
B1: Demonstrates the statement is true for \(n = 1\). Accept as minimum \((1)^2 = 1\) and \(\dfrac{1}{3}(3) = 1\) for the check (both sides must clearly be evaluated as 1), and for the conclusion “true for \(n = 1\)” or broken “if/when \(n = 1\) <check> (hence) true/shown/tick” – must be part of the initial check (not just in the conclusion).
M1: Assume the result for \(n = k\) and attempts to add the \((k + 1)\)th term to the result for the sum to \(k\) terms. An assumption may be clearly made, but accept tacit assumptions. Allow if there are minor slips in the expressions if the intent is clear.
dM1: Makes progress towards proving the inductive step by either factorising the common factor \((2k + 1)\) or expanding fully. There may be variations so score for equivalent progress.
A1: Either for a correct factorised form with \((k + 1)\) as a factor or a fully correct expansion with gathered terms. Must have come from correct work from the inductive step, not backwards worked from the \(n = k + 1\) expression.
A1: Completes the inductive steps by obtaining the correct expression in terms of \((k + 1)\) with suitable intermediate step e.g. clear factorisation of \((k + 1)\) first, OR by expanding the required expression for \(n = k + 1\) to achieve equal expressions. Must have been completely correct work, no errors.
A1: Depends on the MMAA marks having been scored with at least an attempt to check one side of the \(n = 1\) case having been made. Correct complete conclusion. Must include the notions of “true for \(n = 1\)”, “true for \(n = k\) implies true for \(n = k + 1\)” and “hence true for all \(n\)” though the exact wording will vary. The conclusion should be given at the end with the exception that the “true for \(n = 1\)” may be stated with the initial check.
Note: stating “true for \(n = k\) and \(n = k + 1\)” will be A0.
Note: Accept use of \(n\) instead of \(k\) for all except the final A mark.
Note: Attempts at using summation formulae without induction score no marks.