| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Sum/difference of two binomials simplification |
| Difficulty | Easy -1.8 This is a highly routine, scaffolded binomial expansion question requiring only direct application of the binomial theorem to small integer powers (n=3,4), followed by simple algebraic substitution and combination. Parts (a) and (b) are essentially textbook exercises with no problem-solving element, and part (c) merely asks students to subtract the results and observe cancellation of constant and linear terms—purely mechanical work with no conceptual challenge. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((1+x)^3 = 1+3x+3x^2+x^3\) | M1, A1 | Any valid method to expand \((1+x)^3\) fully — Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((1+x)^4 = 1+4x+6x^2+4x^3+x^4\) | M1, A1 | Any valid method to expand \((1+x)^4\) fully — Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((1+4x)^3 = 1+3(4x)+3(4x)^2+(4x)^3\) | M1 | |
| \((1+4x)^3 = 1+12x+48x^2+64x^3\) | A1\(\checkmark\) | Ft on one numerical slip in (a)(i) — Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((1+3x)^4 = 1+4(3x)+6(3x)^2+4(3x)^3+(3x)^4\) | M1 | |
| \(= 1+12x+54x^2+108x^3+81x^4\) | A1\(\checkmark\) | Ft on one numerical slip in (a)(ii) — Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((1+3x)^4-(1+4x)^3 = 1+12x+54x^2+108x^3+81x^4-(1+12x+48x^2+64x^3)\) | M1 | Subtracts the answers to (b) with correct number of terms and combines at least two pairs of like terms |
| \(= 6x^2+44x^3+81x^4\) | A1 | CAO. SC: If no attempt in (b) but full expansions given in working for (c), mark retrospectively — Total: 2 |
## Question 6:
### Part (a)(i)
| Working | Mark | Guidance |
|---------|------|----------|
| $(1+x)^3 = 1+3x+3x^2+x^3$ | M1, A1 | Any valid method to expand $(1+x)^3$ fully — **Total: 2** |
### Part (a)(ii)
| Working | Mark | Guidance |
|---------|------|----------|
| $(1+x)^4 = 1+4x+6x^2+4x^3+x^4$ | M1, A1 | Any valid method to expand $(1+x)^4$ fully — **Total: 2** |
### Part (b)(i)
| Working | Mark | Guidance |
|---------|------|----------|
| $(1+4x)^3 = 1+3(4x)+3(4x)^2+(4x)^3$ | M1 | |
| $(1+4x)^3 = 1+12x+48x^2+64x^3$ | A1$\checkmark$ | Ft on one numerical slip in (a)(i) — **Total: 2** |
### Part (b)(ii)
| Working | Mark | Guidance |
|---------|------|----------|
| $(1+3x)^4 = 1+4(3x)+6(3x)^2+4(3x)^3+(3x)^4$ | M1 | |
| $= 1+12x+54x^2+108x^3+81x^4$ | A1$\checkmark$ | Ft on one numerical slip in (a)(ii) — **Total: 2** |
### Part (c)
| Working | Mark | Guidance |
|---------|------|----------|
| $(1+3x)^4-(1+4x)^3 = 1+12x+54x^2+108x^3+81x^4-(1+12x+48x^2+64x^3)$ | M1 | Subtracts the answers to (b) with correct number of terms and combines at least two pairs of like terms |
| $= 6x^2+44x^3+81x^4$ | A1 | CAO. **SC**: If no attempt in (b) but full expansions given in working for (c), mark retrospectively — **Total: 2** |
---6
\begin{enumerate}[label=(\alph*)]
\item Using the binomial expansion, or otherwise:
\begin{enumerate}[label=(\roman*)]
\item express $( 1 + x ) ^ { 3 }$ in ascending powers of $x$;
\item express $( 1 + x ) ^ { 4 }$ in ascending powers of $x$.
\end{enumerate}\item Hence, or otherwise:
\begin{enumerate}[label=(\roman*)]
\item express $( 1 + 4 x ) ^ { 3 }$ in ascending powers of $x$;
\item express $( 1 + 3 x ) ^ { 4 }$ in ascending powers of $x$.
\end{enumerate}\item Show that the expansion of
$$( 1 + 3 x ) ^ { 4 } - ( 1 + 4 x ) ^ { 3 }$$
can be written in the form
$$p x ^ { 2 } + q x ^ { 3 } + r x ^ { 4 }$$
where $p , q$ and $r$ are integers.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2008 Q6 [10]}}