| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Numerical approximation using expansion |
| Difficulty | Moderate -0.8 This is a straightforward binomial expansion question requiring direct application of the formula with simple arithmetic, followed by a routine substitution interpretation. Part (a) is mechanical calculation with no problem-solving, and part (b) tests basic understanding of how to use the expansion. Below average difficulty for A-level, though not trivial due to fractional coefficient handling. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks |
|---|---|
| 7(a) | (cid:167) x(cid:183) 7 (cid:167)7(cid:183) (cid:167) x(cid:183) (cid:167)7(cid:183) (cid:167) x(cid:183) 2 |
| Answer | Marks | Guidance |
|---|---|---|
| (cid:169) 2(cid:185) (cid:169)1(cid:185) (cid:169) 2(cid:185) (cid:169)2(cid:185) (cid:169) 2(cid:185) | M1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| (cid:169) 2(cid:185) | B1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| (cid:169) 2(cid:185) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| (cid:169) 2(cid:185) | A1 | 1.1b |
| Answer | Marks |
|---|---|
| (b) | (cid:167) x(cid:183) |
| Answer | Marks | Guidance |
|---|---|---|
| substituted for x into the expansion | B1 | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Scheme | Marks |
Question 7:
--- 7(a) ---
7(a) | (cid:167) x(cid:183) 7 (cid:167)7(cid:183) (cid:167) x(cid:183) (cid:167)7(cid:183) (cid:167) x(cid:183) 2
2(cid:16) (cid:32)27 (cid:14) 26. (cid:16) (cid:14) 25. (cid:16) (cid:14)...
(cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184)
(cid:169) 2(cid:185) (cid:169)1(cid:185) (cid:169) 2(cid:185) (cid:169)2(cid:185) (cid:169) 2(cid:185) | M1 | 1.1b
7
(cid:167) x(cid:183)
2(cid:16) (cid:32) 128(cid:14) ...
(cid:168) (cid:184)
(cid:169) 2(cid:185) | B1 | 1.1b
7
(cid:167) x(cid:183)
2(cid:16) (cid:32) ... (cid:16)224x (cid:14)...
(cid:168) (cid:184)
(cid:169) 2(cid:185) | A1 | 1.1b
7
(cid:167) x(cid:183)
2(cid:16) (cid:32) ... (cid:14) ... (cid:14) 168x2 ((cid:14) ...)
(cid:168) (cid:184)
(cid:169) 2(cid:185) | A1 | 1.1b
(4)
(b) | (cid:167) x(cid:183)
Solve 2(cid:16) (cid:32) 1.995 so x = 0.01 and state that 0.01 would be
(cid:168) (cid:184)
(cid:169) 2(cid:185)
substituted for x into the expansion | B1 | 2.4
(1)
(5 marks)
Notes:
(a)
M1: Need correct binomial coefficient with correct power of 2 and correct power of x.
Coefficients may be given in any correct form; e.g. 1, 7, 21 or 7C , 7C , 7C or equivalent
0 1 2
B1: Correct answer, simplified as given in the scheme
A1: Correct answer, simplified as given in the scheme
A1: Correct answer, simplified as given in the scheme
(b)
B1: Needs a full explanation i.e. to state x = 0.01 and that this would be substituted and that it
(cid:167) x(cid:183)
is a solution of 2(cid:16) (cid:32) 1.995
(cid:168) (cid:184)
(cid:169) 2(cid:185)
Question | Scheme | Marks | AOs\begin{enumerate}[label=(\alph*)]
\item Find the first $3$ terms, in ascending powers of $x$, of the binomial expansion of
$$\left(2 - \frac{x}{2}\right)^7$$
giving each term in its simplest form.
[4]
\item Explain how you would use your expansion to give an estimate for the value of $1.995^7$
[1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 Q7 [5]}}