| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2020 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Product with reciprocal term binomial |
| Difficulty | Moderate -0.3 Part (a) is straightforward application of binomial theorem with positive integer power requiring routine calculation of coefficients. Part (b) requires identifying which terms multiply to give constant term, involving algebraic manipulation but following a standard pattern for this question type. More mechanical than conceptual, slightly easier than average A-level question. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(2-\frac{x}{4}\right)^{10} = 2^{10} + \binom{10}{1}2^9\left(-\frac{x}{4}\right) + \binom{10}{2}2^8\left(-\frac{x}{4}\right)^2 + \binom{10}{3}2^7\left(-\frac{x}{4}\right)^3 + ...\) | M1 | Attempts binomial expansion to get third and/or fourth term with acceptable structure. Correct binomial coefficient combined with correct power of \(\frac{x}{4}\) and correct power of 2. Allow \(\pm\binom{10}{2}2^8\left(\pm\frac{x}{4}\right)^2\) or \(\pm{}^{10}C_3 2^7\left(\pm\frac{x}{4}\right)^3\). NB \({}^{10}C_2=45\), \({}^{10}C_3=120\) |
| \(1024 - 1280x\) | B1 | |
| \(720x^2\) or \(-240x^3\) | A1 | |
| \(720x^2\) and \(-240x^3\) | A1 | If any "–"s are "+–"s penalise once on first occurrence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(3-\frac{1}{x}\right)^2 = 9 - \frac{6}{x} + \frac{1}{x^2}\) or \(9-\frac{3}{x}-\frac{3}{x}+\frac{1}{x^2}\) | B1 | Correct expansion. May be implied by work to find constant |
| constant term \(= 9\times1024 - \frac{6}{x}(-1280x) + \frac{1}{x^2}(720x^2)\) | M1 | Depends on having expression of form \(A+\frac{B}{x}+\frac{C}{x^2}\) for \(\left(3-\frac{1}{x}\right)^2\) and at least 3-term quadratic expression from (a). \(A\times\)"\(1024\)"\(+B\times\)"\(-1280\)"\(+C\times\)"\(720\)", \(A,B,C\) non-zero |
| \(= 17616\) | A1 | Correct value. Must be "extracted" if complete expansion shown. True value: \(9216+7680+720\) |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(2-\frac{x}{4}\right)^{10} = 2^{10} + \binom{10}{1}2^9\left(-\frac{x}{4}\right) + \binom{10}{2}2^8\left(-\frac{x}{4}\right)^2 + \binom{10}{3}2^7\left(-\frac{x}{4}\right)^3 + ...$ | M1 | Attempts binomial expansion to get third and/or fourth term with acceptable structure. Correct binomial coefficient combined with correct power of $\frac{x}{4}$ and correct power of 2. Allow $\pm\binom{10}{2}2^8\left(\pm\frac{x}{4}\right)^2$ or $\pm{}^{10}C_3 2^7\left(\pm\frac{x}{4}\right)^3$. NB ${}^{10}C_2=45$, ${}^{10}C_3=120$ |
| $1024 - 1280x$ | B1 | |
| $720x^2$ **or** $-240x^3$ | A1 | |
| $720x^2$ **and** $-240x^3$ | A1 | If any "–"s are "+–"s penalise once on first occurrence |
**(4 marks)**
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(3-\frac{1}{x}\right)^2 = 9 - \frac{6}{x} + \frac{1}{x^2}$ or $9-\frac{3}{x}-\frac{3}{x}+\frac{1}{x^2}$ | B1 | Correct expansion. May be implied by work to find constant |
| constant term $= 9\times1024 - \frac{6}{x}(-1280x) + \frac{1}{x^2}(720x^2)$ | M1 | Depends on having expression of form $A+\frac{B}{x}+\frac{C}{x^2}$ for $\left(3-\frac{1}{x}\right)^2$ and at least 3-term quadratic expression from (a). $A\times$"$1024$"$+B\times$"$-1280$"$+C\times$"$720$", $A,B,C$ non-zero |
| $= 17616$ | A1 | Correct value. Must be "extracted" if complete expansion shown. True value: $9216+7680+720$ |
**(3 marks) — Total 7**
---\begin{enumerate}
\item (a) Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of
\end{enumerate}
$$\left( 2 - \frac { x } { 4 } \right) ^ { 10 }$$
giving each term in its simplest form.\\
(b) Hence find the constant term in the series expansion of
$$\left( 3 - \frac { 1 } { x } \right) ^ { 2 } \left( 2 - \frac { x } { 4 } \right) ^ { 10 }$$
\hfill \mbox{\textit{Edexcel P2 2020 Q1 [7]}}