| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Arithmetic/geometric progression coefficients |
| Difficulty | Standard +0.8 This question requires binomial expansion with a parameter k, then uses the AP condition to form and solve a quadratic equation. While the expansion itself is routine P2 content, the algebraic manipulation to establish the AP relationship and solve for k requires careful multi-step reasoning beyond standard textbook exercises. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(2+\frac{kx}{8}\right)^7 = 2^7 + \binom{7}{1}2^6\!\left(\frac{kx}{8}\right) + \binom{7}{2}2^5\!\left(\frac{kx}{8}\right)^2 + \binom{7}{3}2^4\!\left(\frac{kx}{8}\right)^3 + ...\) | M1 | Attempts binomial expansion up to at least 3rd term with acceptable structure for 3rd or 4th term; correct binomial coefficient with correct powers; M0 for descending powers |
| \(= 128 + 56kx\) | B1 | \(128 + 56kx\); may be listed but must be simplified |
| \(+\, \frac{21}{2}k^2x^2\) | A1 | \(\frac{21}{2}k^2x^2\) or \(\frac{35}{32}k^3x^3\); may be listed, does not need to be simplified, but binomial coefficients must be numerical |
| \(+\, \frac{35}{32}k^3x^3 + ...\) | A1 | Both \(\frac{21}{2}k^2x^2\) and \(\frac{35}{32}k^3x^3\) correct (or exact simplified equivalents); must have \(k^nx^n\) terms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{35}{32}k^3 - \frac{21}{2}k^2 = \frac{21}{2}k^2 - 56k\) | M1 | Sets up equation equating coefficients correctly |
| \(5k^2 - 96k + 256 = 0 \Rightarrow k = ...\) | dM1 | Forms and solves quadratic |
| \(k = 16,\, \frac{16}{5}\) | A1 | Both values |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(2+\frac{kx}{8}\right)^7 = 2^7 + \binom{7}{1}2^6\!\left(\frac{kx}{8}\right) + \binom{7}{2}2^5\!\left(\frac{kx}{8}\right)^2 + \binom{7}{3}2^4\!\left(\frac{kx}{8}\right)^3 + ...$ | M1 | Attempts binomial expansion up to at least 3rd term with acceptable structure for 3rd or 4th term; correct binomial coefficient with correct powers; M0 for descending powers |
| $= 128 + 56kx$ | B1 | $128 + 56kx$; may be listed but must be simplified |
| $+\, \frac{21}{2}k^2x^2$ | A1 | $\frac{21}{2}k^2x^2$ or $\frac{35}{32}k^3x^3$; may be listed, does not need to be simplified, but binomial coefficients must be numerical |
| $+\, \frac{35}{32}k^3x^3 + ...$ | A1 | Both $\frac{21}{2}k^2x^2$ and $\frac{35}{32}k^3x^3$ correct (or exact simplified equivalents); must have $k^nx^n$ terms |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{35}{32}k^3 - \frac{21}{2}k^2 = \frac{21}{2}k^2 - 56k$ | M1 | Sets up equation equating coefficients correctly |
| $5k^2 - 96k + 256 = 0 \Rightarrow k = ...$ | dM1 | Forms and solves quadratic |
| $k = 16,\, \frac{16}{5}$ | A1 | Both values |\begin{enumerate}
\item $\mathrm { f } ( x ) = \left( 2 + \frac { k x } { 8 } \right) ^ { 7 }$ where $k$ is a non-zero constant\\
(a) Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of $\mathrm { f } ( x )$. Give each term in simplest form.
\end{enumerate}
Given that, in the binomial expansion of $\mathrm { f } ( x )$, the coefficients of $x , x ^ { 2 }$ and $x ^ { 3 }$ are the first 3 terms of an arithmetic progression,\\
(b) find, using algebra, the possible values of $k$.\\
(Solutions relying entirely on calculator technology are not acceptable.)
\hfill \mbox{\textit{Edexcel P2 2023 Q3 [7]}}