| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Numerical approximation using expansion |
| Difficulty | Moderate -0.3 This is a straightforward application of the binomial theorem with a standard numerical approximation follow-up. Part (a) requires routine expansion and simplification, part (b) involves recognizing x=1 substitution, and part (c) tests understanding that truncating a series with positive terms gives an underestimate. Slightly easier than average due to being a well-practiced exam technique with clear steps. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(2+\frac{x}{8}\right)^{13} = 8192 + \ldots\) | B1 | Correct constant term of 8192 |
| \(+\binom{13}{1}2^{12}\left(\frac{x}{8}\right)^1 + \binom{13}{2}2^{11}\left(\frac{x}{8}\right)^2 + \binom{13}{3}2^{10}\left(\frac{x}{8}\right)^3 + \ldots\) | M1 | Correct binomial coefficients linked with correct powers of \(x\) at least twice in \(x\), \(x^2\) and \(x^3\) terms |
| \(\left(2+\frac{x}{8}\right)^{13} = (8192) + 6656x + 2496x^2 + 572x^3 + \ldots\) | A1A1 | A1: any two of final three terms correct; A1: all three correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{x}{8} = 0.0125 \Rightarrow x = 0.1\) | B1 | Identifies \(x=0.1\) needs to be substituted |
| \(2.0125^{13} \approx 8192 + 665.6 + 24.96 + 0.572\) | M1 | Substitutes \(x=0.1\) into answer to (a) |
| \(= 8883.132\) cao | A1 | Correct answer, must be to 3 dp |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| As all terms in the expansion are positive, the truncated series will give an underestimate of the actual value | B1 | Must refer to terms being positive; do not accept just "adding more terms" without reference to terms being non-negative |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(2+\frac{x}{8}\right)^{13} = 8192 + \ldots$ | B1 | Correct constant term of 8192 |
| $+\binom{13}{1}2^{12}\left(\frac{x}{8}\right)^1 + \binom{13}{2}2^{11}\left(\frac{x}{8}\right)^2 + \binom{13}{3}2^{10}\left(\frac{x}{8}\right)^3 + \ldots$ | M1 | Correct binomial coefficients linked with correct powers of $x$ at least twice in $x$, $x^2$ and $x^3$ terms |
| $\left(2+\frac{x}{8}\right)^{13} = (8192) + 6656x + 2496x^2 + 572x^3 + \ldots$ | A1A1 | A1: any two of final three terms correct; A1: all three correct |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{x}{8} = 0.0125 \Rightarrow x = 0.1$ | B1 | Identifies $x=0.1$ needs to be substituted |
| $2.0125^{13} \approx 8192 + 665.6 + 24.96 + 0.572$ | M1 | Substitutes $x=0.1$ into answer to (a) |
| $= 8883.132$ cao | A1 | Correct answer, must be to 3 dp |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| As all terms in the expansion are positive, the truncated series will give an underestimate of the actual value | B1 | Must refer to terms being positive; do not accept just "adding more terms" without reference to terms being non-negative |
---\begin{enumerate}
\item (a) Find, in ascending powers of $x$, up to and including the term in $x ^ { 3 }$, the binomial expansion of
\end{enumerate}
$$\left( 2 + \frac { x } { 8 } \right) ^ { 13 }$$
fully simplifying each coefficient.\\
(b) Use the answer to part (a) to find an approximation for $2.0125 ^ { 13 }$
Give your answer to 3 decimal places.
Without calculating $2.0125 { } ^ { 13 }$\\
(c) state, with a reason, whether the answer to part (b) is an overestimate or an underestimate.\\
\hfill \mbox{\textit{Edexcel P2 2021 Q4 [8]}}