A2 June 2023 Q3
3. In a model for the number of subscribers to a new social media channel it is assumed that
- each week 20% of the subscribers at the start of the week cancel their subscriptions
- between the start and end of week \(n\) the channel gains \(20n\) new subscribers
Given that at the end of week 1 there were 25 subscribers,
Given that 6 months after starting the channel there were approximately 1800 subscribers,
| Scheme | Marks | AO |
|---|---|---|
Two of
| M1 | 3.3 |
| All three points above put together in conclusion Hence \(U_{n+1} = 0.8U_n + 20(n + 1),\ U_1 = 25\) | A1 | 2.4 |
| (2) |
Notes
M1: See scheme. Explains how the assumptions lead to at least two of the aspects indicated. Accept less formal explanations as long as the intent is clear.
A1: All three aspects explained and put together to set up the model.
| Scheme | Marks | AO |
|---|---|---|
| \(n = 1 \Rightarrow U_1 = 325 \times 1 + 100 \times 1 - 400 = 325 - 300 = 25\) {so the result is true for \(n = 1\)} | B1 | 2.2a |
| (Assume true for \(n = k\) then)\[U_{k+1} = 0.8U_k + 20k + 20 = 0.8\left(325\left(\frac{4}{5}\right)^{k-1} + 100k - 400\right) + 20k + 20\] | M1 | 1.1b |
| \[= 325 \times \frac{4}{5} \times \left(\frac{4}{5}\right)^{k-1} + 80k - 320 + 20k + 20\]\[= 325 \times \left(\frac{4}{5}\right)^{k} + 100k - 300\] | M1 | 1.1b |
| \[= 325 \times \left(\frac{4}{5}\right)^{k} + 100(k + 1) - 400\] | A1 | 2.1 |
| Hence if the result is true for \(n = k\), then it is true for \(n = k + 1\), and as it is true for \(n = 1\), so it is true for all positive integers \(n\) or \(n \geqslant 1\) | A1 | 2.4 |
| (5) |
Notes
B1: Checks the case for \(n = 1\) holds.
M1: Makes the inductive assumption (may be implied by working) and substitutes the closed form for \(U_k\) into the recurrence relation for \(U_{k+1}\) or equivalent work with different variable (e.g. \(n\) instead of \(k\)) or indexing (e.g. from \(k - 1\) to \(k\)).
M1: Simplifies to the point of combining the powers of \(\dfrac{4}{5}\) to one term.
A1: For correct work leading to the form shown. The \(k + 1\) must be seen in the added term but allow just \(k\) for the power.
A1: For a completely correct proof (all previous marks must be gained) with a conclusion that includes all of the bold statements in the scheme or equivalents.
Alternative
| Scheme | Marks | AO |
|---|---|---|
| \(n = 1 \Rightarrow U_1 = 325 \times 1 + 100 \times 1 - 400 = 325 - 300 = 25\) so the result is true for \(n = 1\) | B1 | 2.2a |
| \[U_{k+1} = 325\left(\frac{4}{5}\right)^{k} + 100(k + 1) - 400 = \frac{4}{5} \times 325\left(\frac{4}{5}\right)^{k-1} + 100k - 300\] | M1 | 1.1b |
| \[= \frac{4}{5} \times \left(325\left(\frac{4}{5}\right)^{k-1} + 100k - 400\right) + 20k + 20\] | M1 | 1.1b |
| \(U_{k+1} = 0.8U_k + 20(k + 1)\) | A1 | 2.1 |
| Hence if the result is true for \(n = k\), then it is true for \(n = k + 1\), and as it is true for \(n = 1\), so it is true for all positive integers \(n\) or \(n \geqslant 1\) | A1 | 2.4 |
| (5) |
B1: Checks the case for \(n = 1\) holds.
M1: Writes out the term \(U_{k+1}\) and starts the process to write in terms of \(U_k\) by factorising out \(\dfrac{4}{5}\) from the first term.
M1: Factorises out \(\dfrac{4}{5}\) to form \(= \dfrac{4}{5} \times \left(325\left(\dfrac{4}{5}\right)^{k-1} + 100k - 400\right) + \ldots\)
A1: For correct work leading \(U_{k+1} = 0.8U_k + 20(k + 1)\)
A1: For a completely correct proof (all previous marks must be gained) with a conclusion that includes all of the bold statements in the scheme or equivalents.
| Scheme | Marks | AO |
|---|---|---|
| An attempt at either \(U_{24} = 325 \times 0.8^{23} + 2400 - 400 = 2001.9\ldots\) Or \(U_{25} = 325 \times 0.8^{24} + 2500 - 400 = 2101.5\ldots\) or \(U_{26} = 325 \times 0.8^{25} + 2600 - 400 = 2201.2\ldots\) Or \(U_{27} = 325 \times 0.8^{26} + 2700 - 400 = 2300.98\ldots\) | M1 | 3.4 |
| Correct value for their number of weeks 24, 25, 26 or 27. Compares the value after 6 months with 1800 and draws a conclusion e.g. This is overestimating the actual amount by 400 people and therefore not a very good model. | A1 | 3.5a |
| (2) | ||
| (9 marks) |
Notes
M1: Evaluates \(U_{24}\) \(U_{25}\) \(U_{26}\) or \(U_{27}\). Allow attempts that deduce \(\left(\dfrac{4}{5}\right)^{n} \to 0\) and just evaluate \(100 \times 26 - 400 = 2200\)
A1: Correct value and appropriate conclusion.