| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Recurrence Relations with Given Sum Condition |
| Difficulty | Standard +0.8 This question requires substituting into a recurrence relation, algebraic manipulation to find expressions for multiple terms, then solving a cubic equation arising from the sum condition. While the individual steps are standard A-level techniques, the combination of recurrence relations with a sum constraint and the resulting algebraic complexity (likely a cubic to solve) elevates this above typical questions. It's more challenging than standard recurrence relation exercises but doesn't require novel mathematical insight. |
| Spec | 1.04e Sequences: nth term and recurrence relations |
| VIXV SIHIANI III IM IONOO | VIAV SIHI NI JYHAM ION OO | VI4V SIHI NI JLIYM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a_2 = 2p^2 - 7\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a_3 = 2(2p^2 - 7 + 3)^2 - 7\) | M1 | Attempt to find \(a_3\) by substituting \(a_2\) into formula. Look for \(2(\ldots)^2 - 7\) where \(\ldots\) is recognisable attempt at \(a_2 + 3\). M0 if \(-7\) is missing |
| \(p - 3 + 2p^2 - 7 + 8p^4 - 32p^2 + 25 = p + 15\) | M1 | Attempts to add first 3 terms and equate to \(p+15\). Not dependent; attempts at three terms may be poor |
| \(\Rightarrow 8p^4 - 30p^2 = 0\) | A1 | Terms gathered according to power (need not all be on one side) |
| Correct values: \(p^2 = 0\) and \(\frac{15}{4}\) (may be implied by \(p=0\) and \(\pm\sqrt{\frac{15}{4}}\)) | A1 | Accept equivalent fractions or decimal answer for \(\frac{15}{4}\), or awrt 1.94 for \(p\). May be called by another letter |
| Uses their values of \(p\) to find \(a_2\) using a correct method | dM1 | Depends on second M mark. Attempts at least one value for \(a_2\) using \(p\) from attempt at solving equation. Must use formula from part (a) |
| Possible values for \(a_2\) are \(-7\) and \(\frac{1}{2}\) | A1cso | Identifies both \(-7\) and \(\frac{1}{2}\) as possible values and no others |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum_{n=1}^{3} a_n = p+15 \Rightarrow p - 3 + a_2 + 2(a_2+3)^2 - 7 = p+15\) | M1 | Attempts sum of first three terms with \(a_3\) in terms of \(a_2\), sets equal to \(p+15\) |
| \(\Rightarrow a_2 + 2(a_2+3)^2 - 25 = 0\) | M1A1 | Cancels \(p\)'s to reach equation in just \(a_2\); correct unsimplified equation |
| \(\Rightarrow 2a_2^2 + 13a_2 - 7 = 0\) | A1 | Correct simplified quadratic in \(a_2\) |
| \(\Rightarrow (2a_2-1)(a_2+7) = 0 \Rightarrow a_2 = \ldots\) | dM1 | Solves the quadratic in \(a_2\) |
| Possible values for \(a_2\) are \(-7\) and \(\frac{1}{2}\) | A1cso | Correct values for \(a_2\) |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a_2 = 2p^2 - 7$ | B1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a_3 = 2(2p^2 - 7 + 3)^2 - 7$ | M1 | Attempt to find $a_3$ by substituting $a_2$ into formula. Look for $2(\ldots)^2 - 7$ where $\ldots$ is recognisable attempt at $a_2 + 3$. M0 if $-7$ is missing |
| $p - 3 + 2p^2 - 7 + 8p^4 - 32p^2 + 25 = p + 15$ | M1 | Attempts to add first 3 terms and equate to $p+15$. Not dependent; attempts at three terms may be poor |
| $\Rightarrow 8p^4 - 30p^2 = 0$ | A1 | Terms gathered according to power (need not all be on one side) |
| Correct values: $p^2 = 0$ and $\frac{15}{4}$ (may be implied by $p=0$ and $\pm\sqrt{\frac{15}{4}}$) | A1 | Accept equivalent fractions or decimal answer for $\frac{15}{4}$, or awrt 1.94 for $p$. May be called by another letter |
| Uses their values of $p$ to find $a_2$ using a correct method | dM1 | Depends on second M mark. Attempts at least one value for $a_2$ using $p$ from attempt at solving equation. Must use formula from part (a) |
| Possible values for $a_2$ are $-7$ and $\frac{1}{2}$ | A1cso | Identifies both $-7$ and $\frac{1}{2}$ as possible values and no others |
### Part (b) Alternative:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{n=1}^{3} a_n = p+15 \Rightarrow p - 3 + a_2 + 2(a_2+3)^2 - 7 = p+15$ | M1 | Attempts sum of first three terms with $a_3$ in terms of $a_2$, sets equal to $p+15$ |
| $\Rightarrow a_2 + 2(a_2+3)^2 - 25 = 0$ | M1A1 | Cancels $p$'s to reach equation in just $a_2$; correct unsimplified equation |
| $\Rightarrow 2a_2^2 + 13a_2 - 7 = 0$ | A1 | Correct simplified quadratic in $a_2$ |
| $\Rightarrow (2a_2-1)(a_2+7) = 0 \Rightarrow a_2 = \ldots$ | dM1 | Solves the quadratic in $a_2$ |
| Possible values for $a_2$ are $-7$ and $\frac{1}{2}$ | A1cso | Correct values for $a_2$ |
---8. A sequence $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots$ is defined by
$$\begin{gathered}
a _ { n + 1 } = 2 \left( a _ { n } + 3 \right) ^ { 2 } - 7 \\
a _ { 1 } = p - 3
\end{gathered}$$
where $p$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $a _ { 2 }$ in terms of $p$, giving your answer in simplest form.
Given that $\sum _ { n = 1 } ^ { 3 } a _ { n } = p + 15$
\item find the possible values of $a _ { 2 }$\\
\begin{center}
\begin{tabular}{|l|l|l|}
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VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
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\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2021 Q8 [7]}}