Moderate -0.8 This is a straightforward C2 binomial expansion question requiring routine application of the binomial theorem formula to find the first 4 terms, followed by solving a simple quadratic equation. The mechanics are standard with no conceptual challenges or problem-solving insight required.
4. (a) Find the first 4 terms, in ascending powers of \(x\), of the binomial expansion of \(( 1 + a x ) ^ { 7 }\), where \(a\) is a constant. Give each term in its simplest form.
Given that the coefficient of \(x ^ { 2 }\) in this expansion is 525 ,
(b) find the possible values of \(a\).
Evidence from one of these terms is enough. \(\binom{7}{2}\) and \(\binom{7}{3}\) acceptable, but not \(\left(\frac{7}{2}\right)\) or \(\left(\frac{7}{3}\right)\) unless corrected
\(+21a^2x^2\)
A1
or \(+21(ax)^2\) or \(+21(a^2x^2)\). Binomial coefficients must be simplified
\(+35a^3x^3\)
A1
or \(+35(ax)^3\) or \(+35(a^3x^3)\)
Part (b):
Answer
Marks
Guidance
Answer
Marks
Guidance
\(21a^2 = 525\)
M1
Equating coefficient of \(x^2\) to 525. An equation in \(a\) or \(a^2\) alone required. \(21a^2x^2=525 \Rightarrow 21a^2=525\) acceptable, but \(21a^2x^2=525 \Rightarrow a^2=25\) is not
\(a = \pm 5\) (both values required)
A1
Answer \(a=5\) with no working scores M1 A0
## Question 4:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(1+ax)^7 = 1 + 7ax\ldots$ | B1 | Not unsimplified versions |
| $+\frac{7\times6}{2}(ax)^2 + \frac{7\times6\times5}{6}(ax)^3$ | M1 | Evidence from one of these terms is enough. $\binom{7}{2}$ and $\binom{7}{3}$ acceptable, but not $\left(\frac{7}{2}\right)$ or $\left(\frac{7}{3}\right)$ unless corrected |
| $+21a^2x^2$ | A1 | or $+21(ax)^2$ or $+21(a^2x^2)$. Binomial coefficients must be simplified |
| $+35a^3x^3$ | A1 | or $+35(ax)^3$ or $+35(a^3x^3)$ |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $21a^2 = 525$ | M1 | Equating coefficient of $x^2$ to 525. An equation in $a$ or $a^2$ alone required. $21a^2x^2=525 \Rightarrow 21a^2=525$ acceptable, but $21a^2x^2=525 \Rightarrow a^2=25$ is not |
| $a = \pm 5$ (both values required) | A1 | Answer $a=5$ with no working scores M1 A0 |
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4. (a) Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of $( 1 + a x ) ^ { 7 }$, where $a$ is a constant. Give each term in its simplest form.
Given that the coefficient of $x ^ { 2 }$ in this expansion is 525 ,\\
(b) find the possible values of $a$.\\
\hfill \mbox{\textit{Edexcel C2 2010 Q4 [6]}}