| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Standard product of two binomials |
| Difficulty | Moderate -0.3 Part (a) is straightforward application of binomial theorem with positive integer power requiring calculation of three coefficients using nCr. Part (b) requires multiplying the result by (1-x/2)^2 and collecting x^3 terms—a standard 'hence' extension that adds one extra step but remains routine for C2 level. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((1+2x)^7 = 1 + \binom{7}{1}(2x)^1 + \binom{7}{2}(2x)^2 + \binom{7}{3}(2x)^3 + \ldots\) | M1 | Any valid method. PI by a correct value for either \(a\) or \(b\) or \(c\) |
| \(= 1 + 14x + 84x^2 + 280x^3 + \ldots\) \(\{a = 14,\ b = 84,\ c = 280\}\) | A1 \(\times\) 3 | A1 for each of \(a, b, c\). SC \(a=7\), \(b=21\), \(c=35\) either explicitly or within expn (M1A0) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left(1 - \frac{1}{2}x\right)^2 = 1 - x + \frac{1}{4}x^2\) | B1 | Correct expansion stated explicitly or used later |
| \(x^3\) terms from expansion of \(\left(1-\frac{1}{2}x\right)^2(1+2x)^7\) are \(cx^3\) and \(-x(bx^2)\) and \(\frac{1}{4}x^2(ax)\) | M1 | Any one of the three, or ft on \(c\)'s non-zero values for \(a\), \(b\) or \(c\). Must be from products of terms using \(c\)'s two expansions |
| \(cx^3 - x(bx^2) + \frac{1}{4}x^2(ax)\) | A1F | ft \(c\)'s two expansions provided all three combinations of terms are present |
| Coefficient of \(x^3\) is \(c - b + 0.25a = 199.5\) | A1 | OE eg \(\frac{399}{2}\). Condone \(199.5x^3\) |
## Question 7:
### Part 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(1+2x)^7 = 1 + \binom{7}{1}(2x)^1 + \binom{7}{2}(2x)^2 + \binom{7}{3}(2x)^3 + \ldots$ | M1 | Any valid method. PI by a correct value for either $a$ or $b$ or $c$ |
| $= 1 + 14x + 84x^2 + 280x^3 + \ldots$ $\{a = 14,\ b = 84,\ c = 280\}$ | A1 $\times$ 3 | A1 for each of $a, b, c$. SC $a=7$, $b=21$, $c=35$ either explicitly or within expn (M1A0) |
### Part 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left(1 - \frac{1}{2}x\right)^2 = 1 - x + \frac{1}{4}x^2$ | B1 | Correct expansion stated explicitly or used later |
| $x^3$ terms from expansion of $\left(1-\frac{1}{2}x\right)^2(1+2x)^7$ are $cx^3$ and $-x(bx^2)$ and $\frac{1}{4}x^2(ax)$ | M1 | Any one of the three, or ft on $c$'s non-zero values for $a$, $b$ or $c$. Must be from products of terms using $c$'s two expansions |
| $cx^3 - x(bx^2) + \frac{1}{4}x^2(ax)$ | A1F | ft $c$'s two expansions provided all three combinations of terms are present |
| Coefficient of $x^3$ is $c - b + 0.25a = 199.5$ | A1 | OE eg $\frac{399}{2}$. Condone $199.5x^3$ |
---7
\begin{enumerate}[label=(\alph*)]
\item The first four terms of the binomial expansion of $( 1 + 2 x ) ^ { 7 }$ in ascending powers of $x$ are $1 + a x + b x ^ { 2 } + c x ^ { 3 }$. Find the values of the integers $a , b$ and $c$.
\item Hence find the coefficient of $x ^ { 3 }$ in the expansion of $\left( 1 - \frac { 1 } { 2 } x \right) ^ { 2 } ( 1 + 2 x ) ^ { 7 }$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2010 Q7 [8]}}