| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Find constants from coefficient conditions on terms |
| Difficulty | Moderate -0.8 This is a straightforward binomial expansion question requiring routine application of formulas. Part (a) involves simple coefficient matching to find k and p using standard binomial coefficients. Part (b) requires identifying which terms contribute to x², a mechanical process. The recurrence relation part is basic substitution and solving a linear equation. All steps are standard textbook exercises with no novel insight required. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States/uses either \(16k = -4\) or \(\frac{16 \times 15}{2}k^2 = p\) | M1 | May be implied by a correct value for \(k\) or \(p\) if no incorrect working seen |
| (i) \(k = -\frac{1}{4}\) or \(-0.25\) | A1 | |
| (ii) \(p = \frac{15}{2}\) or \(7.5\) | A1 | Allow if this follows from \(k = \pm\frac{1}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts either \(2"p"\) or \(16 \times \frac{16 \times 15 \times 14}{3!} \times k^3\) | M1 | Either \(2p(x^2)\) or \(16 \times \frac{16\times15\times14}{3!} \times k^3(x^2)\); may be part of full expansion; may be in terms of \(p\) and \(k\) |
| Attempts sum of \(2"p"\) and \(16 \times \frac{16\times15\times14}{3!} \times k^3\) | dM1 | Must have attempted at least one term correctly; must arise from correct combination of powers; coefficients need not be correct; must be numerical terms |
| Term in \(x^2 = (15-140)x^2 = -125x^2\) | A1 | Term in \(-125x^2\); must include the \(x^2\); do not allow as part of an expansion for this mark |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| States/uses either $16k = -4$ or $\frac{16 \times 15}{2}k^2 = p$ | M1 | May be implied by a correct value for $k$ or $p$ if no incorrect working seen |
| (i) $k = -\frac{1}{4}$ or $-0.25$ | A1 | |
| (ii) $p = \frac{15}{2}$ or $7.5$ | A1 | Allow if this follows from $k = \pm\frac{1}{4}$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts either $2"p"$ or $16 \times \frac{16 \times 15 \times 14}{3!} \times k^3$ | M1 | Either $2p(x^2)$ or $16 \times \frac{16\times15\times14}{3!} \times k^3(x^2)$; may be part of full expansion; may be in terms of $p$ and $k$ |
| Attempts sum of $2"p"$ and $16 \times \frac{16\times15\times14}{3!} \times k^3$ | dM1 | Must have attempted at least one term correctly; must arise from correct combination of powers; coefficients need not be correct; must be numerical terms |
| Term in $x^2 = (15-140)x^2 = -125x^2$ | A1 | Term in $-125x^2$; must include the $x^2$; do not allow as part of an expansion for this mark |
---\begin{enumerate}
\item The first three terms, in ascending powers of $x$, of the binomial expansion of $( 1 + k x ) ^ { 16 }$ are
\end{enumerate}
$$1 , - 4 x \text { and } p x ^ { 2 }$$
where $k$ and $p$ are constants.\\
(a) Find, in simplest form,\\
(i) the value of $k$\\
(ii) the value of $p$
$$g ( x ) = \left( 2 + \frac { 16 } { x } \right) ( 1 + k x ) ^ { 16 }$$
Using the value of $k$ found in part (a),\\
(b) find the term in $x ^ { 2 }$ in the expansion of $\mathrm { g } ( x )$.
$$\begin{aligned}
u _ { 1 } & = 6 \\
u _ { n + 1 } & = k u _ { n } + 3
\end{aligned}$$
where $k$ is a positive constant.\\
(a) Find, in terms of $k$, an expression for $u _ { 3 }$
Given that $\sum _ { n = 1 } ^ { 3 } u _ { n } = 117$\\
(b) find the value of $k$.\\
\hfill \mbox{\textit{Edexcel P2 2021 Q1 [6]}}