| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Binomial times linear coefficient |
| Difficulty | Moderate -0.3 This is a standard C2 binomial expansion question with straightforward algebraic manipulation. Part (a) requires routine application of binomial theorem, while parts (b)-(d) involve simple coefficient matching and arithmetic. The multi-part structure adds length but not conceptual difficulty—all steps are mechanical once the expansion is performed. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) \((2-9x)^4 = 2^4 + ^4C_1 2^3(-9x) + ^4C_2 2^2(-9x)^2\) | B1 | First term of 16 in their final series |
| Way 1: At least one of \(\left(^4C_1 \times ... \times x\right)\) or \(\left(^4C_2 \times ... \times x^2\right)\) | M1 | |
| \(= (16) - 288x + 1944x^2\) | A1 | At least one of \(-288x\) or \(+1944x^2\) (allow \(\pm 288x\)) |
| [4] | Both \(-288x\) and \(+1944x^2\) | |
| A1 | ||
| Way 2: \((2-9x)^4 = (4-36x+81x^2)(4-36x+81x^2)\) | First term of 16 in their final series | |
| \(= 16 - 144x + 324x^2 - 144x + 1296x^2 - 324x^3\) | M1 | Attempts to multiply a 3 term quadratic by the same 3 term quadratic to achieve either 2 terms in \(x\) or at least 2 terms in \(x^2\) |
| \(= (16) - 288x + 1944x^2\) | A1 | At least one of \(-288x\) or \(+1944x^2\) |
| Both \(-288x\) and \(+1944x^2\) | ||
| A1 | ||
| [4] | ||
| Way 3: \(\{(2-9x)^4 = \} 2^4\left(1 - \frac{9}{2}x\right)^4\) | B1 | First term of 16 in final series |
| \(= 2^4\left(1 + 4\left(-\frac{9}{2}\right) + \frac{4(3)}{2}\left(-\frac{9}{2}\right)^2 + ...\right)\) | M1 | At least one of \(\left(4 \times ... \times x\right)\) or \(\left(\frac{4(3)}{2} \times ... \times x^2\right)\) |
| \(= (16) - 288x + 1944x^2\) | A1 | At least one of \(-288x\) or \(+1944x^2\) |
| Both \(-288x\) and \(+1944x^2\) | ||
| A1 | ||
| [4] | ||
| (b) Parts (b), (c) and (d) may be marked together. \(A = "16"\) | B1ft | Follow through their value from (a) |
| (c) \(\{(1+kx)(2-9x)^4 = \} (1+kx)\{16-288x + \{1944x^2 + ...\}\}\) | M1 | May be seen in part (b) or (d) and can be implied by work in parts (c) or (d) |
| \(x\) terms: \(-288x + 16kx = -232x\) | ||
| giving, \(16k = 56 \Rightarrow k = \frac{7}{2}\) | A1 | \(k = \frac{7}{2}\) o.c. so 3.5 is acceptable |
| [2] | ||
| (d) \(x^2\) terms: \(1944x^2 - 288kx^2\) | M1 | Multiplies out their \((1+kx)\{16-288x+1944x^2 + ...\}\) to give exactly two terms (or coefficients) in \(x^2\) and attempts to find \(B\) using these two terms and a numerical value of \(k\) |
| So, \(B = 1944 - 288\left(\frac{7}{2}\right) = 1944 - 1008 = 936\) | A1 | 936 |
| [2] |
| Answer | Marks |
|---|---|
| Section | Details |
| Ways 1 and 3 | B1 cao: 16 |
| Way 2b | Special Case: Slight Variation on the solution given in the scheme \((2-9x)^4 = (2-9x)(2-9x)(4-36x+81x^2) = (2-9x)(8-108x+486x^2 + ...) = 16 - 216x + 972x^2 - 72x + 972x^2 = (16) - 288x + 1944x^2 + ...\) First term of 16, Multiplies out to give either 2 terms in \(x\) or 2 terms in \(x^2\), At least one of \(-288x\) or \(+1944x^2\), Both \(-288x\) and \(+1944x^2\) |
| (b) | B1ft: Must identify \(A = 16\) or \(A = \text{their constant term found in part (a)}\). Or may write just 16 if this is clearly their answer to part (b). If they expand their series and have 16 as first term of a series it is not sufficient for this mark. |
| (c) | M1: Candidate shows intention to multiply \((1+kx)\) by part of their series from (a) e.g. Just \((1+kx)(16-288x+...)\) or \((1+kx)\{16-288x+1944x^2+...\}\) as fine for M1. Note: This mark can also be implied by candidate multiplying out to find two terms (or coefficients) in \(x\). i.e. f.t. their \(-288x + 16kx\). N.B. \(-288kx = -232x\) with no evidence of brackets is M0 – allow copying slips, or use of factored series, as this is a method mark. If the candidate then divides their final correct answer through by 8 or any other common factor then isw and mark correct series when first seen. So (a) B1M1A1 .It is likely that this approach will be followed by (b) B0, (c) M1A0, (d) M1A0 if they continue with their new series e.g. \(2 - 36x + 283x^2 + ...\) (Do not ft the value 2 as a mark was awarded for 16) |
| (d) | M1: Multiplies out their \((1+kx)\{16-288x+1944x^2 + ...\}\) to give exactly two terms (or coefficients) in \(x^2\) and attempts to find \(B\) using these two terms and a numerical value of \(k\). A1: 936 Note: Award A0 for \(B = 936x^2\) But allow A1 for \(B = 936x^2\) followed by \(B = 936\) and treat this as a correction Correct answers in parts (c) and (d) with no method shown may be awarded full credit. |
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** $(2-9x)^4 = 2^4 + ^4C_1 2^3(-9x) + ^4C_2 2^2(-9x)^2$ | B1 | First term of 16 in their final series |
| **Way 1:** At least one of $\left(^4C_1 \times ... \times x\right)$ or $\left(^4C_2 \times ... \times x^2\right)$ | M1 | |
| $= (16) - 288x + 1944x^2$ | A1 | At least one of $-288x$ or $+1944x^2$ (allow $\pm 288x$) |
| | [4] | Both $-288x$ and $+1944x^2$ |
| | | A1 |
| **Way 2:** $(2-9x)^4 = (4-36x+81x^2)(4-36x+81x^2)$ | | First term of 16 in their final series |
| $= 16 - 144x + 324x^2 - 144x + 1296x^2 - 324x^3$ | M1 | Attempts to multiply a 3 term quadratic by the same 3 term quadratic to achieve either 2 terms in $x$ or at least 2 terms in $x^2$ |
| $= (16) - 288x + 1944x^2$ | A1 | At least one of $-288x$ or $+1944x^2$ |
| | | Both $-288x$ and $+1944x^2$ |
| | | A1 |
| | [4] | |
| **Way 3:** $\{(2-9x)^4 = \} 2^4\left(1 - \frac{9}{2}x\right)^4$ | B1 | First term of 16 in final series |
| $= 2^4\left(1 + 4\left(-\frac{9}{2}\right) + \frac{4(3)}{2}\left(-\frac{9}{2}\right)^2 + ...\right)$ | M1 | At least one of $\left(4 \times ... \times x\right)$ or $\left(\frac{4(3)}{2} \times ... \times x^2\right)$ |
| $= (16) - 288x + 1944x^2$ | A1 | At least one of $-288x$ or $+1944x^2$ |
| | | Both $-288x$ and $+1944x^2$ |
| | | A1 |
| | [4] | |
| **(b)** Parts (b), (c) and (d) may be marked together. $A = "16"$ | B1ft | Follow through their value from (a) |
| **(c)** $\{(1+kx)(2-9x)^4 = \} (1+kx)\{16-288x + \{1944x^2 + ...\}\}$ | M1 | May be seen in part (b) or (d) and can be implied by work in parts (c) or (d) |
| $x$ terms: $-288x + 16kx = -232x$ | | |
| giving, $16k = 56 \Rightarrow k = \frac{7}{2}$ | A1 | $k = \frac{7}{2}$ o.c. so 3.5 is acceptable |
| | [2] | |
| **(d)** $x^2$ terms: $1944x^2 - 288kx^2$ | M1 | Multiplies out their $(1+kx)\{16-288x+1944x^2 + ...\}$ to give exactly two terms (or coefficients) in $x^2$ and attempts to find $B$ using these two terms and a numerical value of $k$ |
| So, $B = 1944 - 288\left(\frac{7}{2}\right) = 1944 - 1008 = 936$ | A1 | 936 |
| | [2] | |
**Question 5 Notes:**
| Section | Details |
|---|---|
| Ways 1 and 3 | B1 cao: 16 |
| Way 2b | Special Case: Slight Variation on the solution given in the scheme $(2-9x)^4 = (2-9x)(2-9x)(4-36x+81x^2) = (2-9x)(8-108x+486x^2 + ...) = 16 - 216x + 972x^2 - 72x + 972x^2 = (16) - 288x + 1944x^2 + ...$ First term of 16, Multiplies out to give either 2 terms in $x$ or 2 terms in $x^2$, At least one of $-288x$ or $+1944x^2$, Both $-288x$ and $+1944x^2$ |
| (b) | B1ft: Must identify $A = 16$ or $A = \text{their constant term found in part (a)}$. Or may write just 16 if this is clearly their answer to part (b). If they expand their series and have 16 as first term of a series it is not sufficient for this mark. |
| (c) | M1: Candidate shows intention to multiply $(1+kx)$ by part of their series from (a) e.g. Just $(1+kx)(16-288x+...)$ or $(1+kx)\{16-288x+1944x^2+...\}$ as fine for M1. Note: This mark can also be implied by candidate multiplying out to find **two terms** (or coefficients) in $x$. i.e. f.t. their $-288x + 16kx$. N.B. $-288kx = -232x$ with no evidence of brackets is M0 – allow copying slips, or use of factored series, as this is a method mark. If the candidate then divides their final correct answer through by 8 or any other common factor then isw and mark correct series when first seen. So (a) B1M1A1 .It is likely that this approach will be followed by (b) B0, (c) M1A0, (d) M1A0 if they continue with their new series e.g. $2 - 36x + 283x^2 + ...$ (Do not ft the value 2 as a mark was awarded for 16) |
| (d) | M1: Multiplies out their $(1+kx)\{16-288x+1944x^2 + ...\}$ to give **exactly two terms** (or coefficients) in $x^2$ and attempts to find $B$ using these two terms and a numerical value of $k$. A1: 936 Note: Award A0 for $B = 936x^2$ But allow A1 for $B = 936x^2$ followed by $B = 936$ and treat this as a correction Correct answers in parts (c) and (d) with no method shown may be awarded full credit. |
---\begin{enumerate}
\item (a) Find the first 3 terms, in ascending powers of $x$, of the binomial expansion of
\end{enumerate}
$$( 2 - 9 x ) ^ { 4 }$$
giving each term in its simplest form.
$$f ( x ) = ( 1 + k x ) ( 2 - 9 x ) ^ { 4 } , \text { where } k \text { is a constant }$$
The expansion, in ascending powers of $x$, of $\mathrm { f } ( x )$ up to and including the term in $x ^ { 2 }$ is
$$A - 232 x + B x ^ { 2 }$$
where $A$ and $B$ are constants.\\
(b) Write down the value of $A$.\\
(c) Find the value of $k$.\\
(d) Hence find the value of $B$.
\hfill \mbox{\textit{Edexcel C2 2016 Q5 [9]}}