| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Product with reciprocal term binomial |
| Difficulty | Standard +0.3 This is a straightforward binomial expansion question requiring standard application of the formula with fractional coefficients, followed by a product that needs careful term collection. Part (a) is routine; part (b) requires identifying which terms from the expansion multiply to give x², which is a small step beyond basic recall but still a standard exam technique. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| \(3^8\) or \(6561\) as constant term | B1 | Sight of \(3^8\) or \(6561\) |
| \(\left(3-\frac{2x}{9}\right)^8 = \ldots + \binom{8}{1}(3)^7\left(-\frac{2x}{9}\right) + \binom{8}{2}(3)^6\left(-\frac{2x}{9}\right)^2 + \binom{8}{3}(3)^5\left(-\frac{2x}{9}\right)^3 + \ldots\) | M1 | Correct structure of 2nd, 3rd or 4th term with correct binomial coefficient, correct power of 3, correct power of \(\left(\pm\frac{2x}{9}\right)\) |
| \(= \ldots + 8\times(3)^7\left(-\frac{2x}{9}\right) + 28\times(3)^6\left(-\frac{2x}{9}\right)^2 + 56(3)^5\left(-\frac{2x}{9}\right)^3\) | A1 | Correct simplified or unsimplified 2nd or 4th term with binomial coefficients evaluated |
| \(= 6561 - 3888x + 1008x^2 - \frac{448}{3}x^3 + \ldots\) | A1 | All four terms correct; allow exact equivalent to \(-\frac{448}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Coefficient of \(x^2\) is \(\frac{1}{2}\times\text{"1008"} - \frac{1}{2}\times\text{"} - \frac{448}{3}\text{"}\) | M1 | Correct method using coefficients from part (a) |
| \(= \frac{1736}{3}\) (or \(578\frac{2}{3}\)) | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Sight of \(3^8(1+.....)\) or \(6561\) as the constant term | B1 | |
| Attempt at binomial expansion \(\left(1-\frac{2}{27}x\right)^8\); correct structure of 2nd, 3rd or 4th term with correct binomial coefficient associated with correct power of \(\left(\pm\frac{2x}{27}\right)\) | M1 | Condone invisible brackets |
| Score for any of: \(8\times-\frac{2}{27}x\), \(\frac{8\times7}{2}\times\left(-\frac{2}{27}x\right)^2\), \(\frac{8\times7\times6}{6}\times\left(-\frac{2}{27}x\right)^3\) implied by \(-\frac{16}{27}x\), \(+\frac{112}{729}x^2\), \(-\frac{448}{19683}x^3\) | ||
| Correct simplified or unsimplified second or fourth term multiplied by \(3^8\) | A1 | |
| \(6561-3888x+1008x^2-\frac{448}{3}x^3\) | A1 | Ignore extra terms; allow \(-\frac{448}{3}\) eg \(-149.\dot{3}\) but not \(-149.3\); condone \(x^1\) and eg \(+-3888x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct strategy: attempt \(\pm\frac{1}{2}\) their coefficient of \(x^2\) from (a) \(\pm\frac{1}{2}\) their coefficient of \(x^3\) from (a); attempt to combine to single value | M1 | Condone any appearance of \(x^2\) or \(x^3\) in intermediate working |
| \(\frac{1736}{3}\) or \(578\frac{2}{3}\) | A1 | Do not accept \(578.\dot{6}\) or \(\frac{1736}{3}x^2\) |
## Question 6(a):
$3^8$ or $6561$ as constant term | B1 | Sight of $3^8$ or $6561$
$\left(3-\frac{2x}{9}\right)^8 = \ldots + \binom{8}{1}(3)^7\left(-\frac{2x}{9}\right) + \binom{8}{2}(3)^6\left(-\frac{2x}{9}\right)^2 + \binom{8}{3}(3)^5\left(-\frac{2x}{9}\right)^3 + \ldots$ | M1 | Correct structure of 2nd, 3rd or 4th term with correct binomial coefficient, correct power of 3, correct power of $\left(\pm\frac{2x}{9}\right)$
$= \ldots + 8\times(3)^7\left(-\frac{2x}{9}\right) + 28\times(3)^6\left(-\frac{2x}{9}\right)^2 + 56(3)^5\left(-\frac{2x}{9}\right)^3$ | A1 | Correct simplified or unsimplified 2nd or 4th term with binomial coefficients evaluated
$= 6561 - 3888x + 1008x^2 - \frac{448}{3}x^3 + \ldots$ | A1 | All four terms correct; allow exact equivalent to $-\frac{448}{3}$
**Total: 4 marks**
---
## Question 6(b):
Coefficient of $x^2$ is $\frac{1}{2}\times\text{"1008"} - \frac{1}{2}\times\text{"} - \frac{448}{3}\text{"}$ | M1 | Correct method using coefficients from part (a)
$= \frac{1736}{3}$ (or $578\frac{2}{3}$) | A1 | Correct value
**Total: 2 marks**
## Question 6 (Alt a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Sight of $3^8(1+.....)$ or $6561$ as the constant term | B1 | |
| Attempt at binomial expansion $\left(1-\frac{2}{27}x\right)^8$; correct structure of 2nd, 3rd or 4th term with correct binomial coefficient associated with correct power of $\left(\pm\frac{2x}{27}\right)$ | M1 | Condone invisible brackets |
| Score for any of: $8\times-\frac{2}{27}x$, $\frac{8\times7}{2}\times\left(-\frac{2}{27}x\right)^2$, $\frac{8\times7\times6}{6}\times\left(-\frac{2}{27}x\right)^3$ implied by $-\frac{16}{27}x$, $+\frac{112}{729}x^2$, $-\frac{448}{19683}x^3$ | | |
| Correct simplified or unsimplified **second** or **fourth** term multiplied by $3^8$ | A1 | |
| $6561-3888x+1008x^2-\frac{448}{3}x^3$ | A1 | Ignore extra terms; allow $-\frac{448}{3}$ eg $-149.\dot{3}$ but not $-149.3$; condone $x^1$ and eg $+-3888x$ |
**Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct strategy: attempt $\pm\frac{1}{2}$ their coefficient of $x^2$ from (a) $\pm\frac{1}{2}$ their coefficient of $x^3$ from (a); attempt to combine to single value | M1 | Condone any appearance of $x^2$ or $x^3$ in intermediate working |
| $\frac{1736}{3}$ or $578\frac{2}{3}$ | A1 | Do not accept $578.\dot{6}$ or $\frac{1736}{3}x^2$ |
---\begin{enumerate}
\item (a) Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of
\end{enumerate}
$$\left( 3 - \frac { 2 x } { 9 } \right) ^ { 8 }$$
giving each term in simplest form.
$$f ( x ) = \left( \frac { x - 1 } { 2 x } \right) \left( 3 - \frac { 2 x } { 9 } \right) ^ { 8 }$$
(b) Find the coefficient of $x ^ { 2 }$ in the series expansion of $\mathrm { f } ( x )$, giving your answer as a simplified fraction.
\hfill \mbox{\textit{Edexcel AS Paper 1 2022 Q6 [6]}}