Standard +0.3 This is a straightforward application of the binomial expansion for fractional powers with a substitution. Part (a) requires routine use of (1+u)^(1/3) with u=4x³, involving standard coefficient calculations. Part (b) is a simple numerical substitution. The question is slightly easier than average as it explicitly asks for 'non-zero terms' (eliminating the need to identify which terms vanish) and the algebra is cleaner than typical binomial questions due to the x³ term.
2. (a) Find, in ascending powers of \(x\), the first three non-zero terms of the binomial series expansion of
$$\sqrt [ 3 ] { 1 + 4 x ^ { 3 } } \quad | x | < \frac { 1 } { \sqrt [ 3 ] { 4 } }$$
giving each coefficient as a simplified fraction.
(b) Use the expansion from part (a) with \(x = \frac { 1 } { 3 }\) to find a rational approximation to \(\sqrt [ 3 ] { 31 }\)
(3)
Correct structure for third term: correct binomial coefficient combined with correct power of \(4x^3\). Condone missing brackets e.g. \(\frac{\frac{1}{3}(\frac{1}{3}-1)}{2}4x^{3^2}\)
Attempts \(1+4\times\left(\frac{1}{3}\right)^3\) and reaches \(\frac{31}{k}\) (usually seen as \(\sqrt[3]{\frac{31}{27}}\)) or substitutes \(x=\frac{1}{3}\) into expansion from part (a)
Attempts \(3\times\) (their expansion from part (a) with \(x=\frac{1}{3}\) substituted). Allow if more terms included. Must be convinced they attempted the calculation and not just found cube root of 31 on calculator
\(= \frac{6869}{2187}\)
A1
Correct fraction
## Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(1+4x^3)^{\frac{1}{3}} = 1 + \frac{1}{3}(\ldots x^{\ldots}) + \ldots$ | M1 | For $1 \pm \frac{1}{3}(4x^{\ldots})$. Allow if the 4 and/or power of $x$ is incorrect/missing |
| $(1+4x^3)^{\frac{1}{3}} = \ldots + \frac{\frac{1}{3}\left(\frac{-2}{3}\right)}{2}(4x^3)^2 + \ldots$ | M1 | Correct structure for third term: correct binomial coefficient combined with correct power of $4x^3$. Condone missing brackets e.g. $\frac{\frac{1}{3}(\frac{1}{3}-1)}{2}4x^{3^2}$ |
| $= 1 + \frac{4}{3}x^3 + \ldots$ or $= \ldots - \frac{16}{9}x^6$ | A1 | Allow mixed fractions e.g. $1\frac{1}{3}$ for $\frac{4}{3}$ but do not allow $'+-'$ for $'-'$ |
| $(1+4x^3)^{\frac{1}{3}} = 1 + \frac{4}{3}x^3 - \frac{16}{9}x^6 + \ldots$ | A1 | **Ignore any extra terms** |
---
## Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1 + 4x^3 = 1 + 4\times\left(\frac{1}{3}\right)^3 = \frac{31}{\ldots}$ or $1 + \frac{4}{3}\left(\frac{1}{3}\right)^3 - \frac{16}{9}\left(\frac{1}{3}\right)^6$ | M1 | Attempts $1+4\times\left(\frac{1}{3}\right)^3$ and reaches $\frac{31}{k}$ (usually seen as $\sqrt[3]{\frac{31}{27}}$) **or** substitutes $x=\frac{1}{3}$ into expansion from part (a) |
| $\sqrt[3]{31} \approx 3\times\left(1 + \frac{4}{3}\left(\frac{1}{3}\right)^3 - \frac{16}{9}\left(\frac{1}{3}\right)^6\right)$ | dM1 | Attempts $3\times$ (their expansion from part (a) with $x=\frac{1}{3}$ substituted). Allow if more terms included. Must be convinced they attempted the calculation and not just found cube root of 31 on calculator |
| $= \frac{6869}{2187}$ | A1 | Correct fraction |
---
2. (a) Find, in ascending powers of $x$, the first three non-zero terms of the binomial series expansion of
$$\sqrt [ 3 ] { 1 + 4 x ^ { 3 } } \quad | x | < \frac { 1 } { \sqrt [ 3 ] { 4 } }$$
giving each coefficient as a simplified fraction.\\
(b) Use the expansion from part (a) with $x = \frac { 1 } { 3 }$ to find a rational approximation to $\sqrt [ 3 ] { 31 }$\\
(3)
\hfill \mbox{\textit{Edexcel P4 2022 Q2 [7]}}