Moderate -0.3 Part (a) is straightforward application of binomial theorem with positive integer power requiring routine calculation of coefficients. Part (b) requires identifying which terms from the expansion combine to give x^0, involving algebraic manipulation but following a standard method. This is slightly easier than average due to the mechanical nature of both parts, though part (b) requires some care in tracking powers of x.
3. (a) Find the first 4 terms, in ascending powers of \(x\), in the binomial expansion of
$$\left( 1 + \frac { x } { 4 } \right) ^ { 12 }$$
giving each coefficient in its simplest form.
(b) Find the term independent of \(x\) in the expansion of
$$\left( \frac { x ^ { 2 } + 8 } { x ^ { 5 } } \right) \left( 1 + \frac { x } { 4 } \right) ^ { 12 }$$
Correct attempt at binomial expansion; accept sight of \(^{12}C_2\left(\frac{x}{4}\right)^2\) or \(^{12}C_3\left(\frac{x}{4}\right)^3\) condoning omission of brackets. FYI \(^{12}C_2=66\), \(^{12}C_3=220\)
\(= 1+3x+\frac{33}{8}x^2+\frac{55}{16}x^3\)
B1, A1
B1 for \(1+3x\) must be simplified; A1 for \(+\frac{33}{8}x^2+\frac{55}{16}x^3\), accept \(+4.125x^2+3.4375x^3\)
Part (b):
Answer
Marks
Guidance
Answer/Working
Marks
Guidance
Sight of term independent of \(x\): \("\frac{55}{16}"\) or \(8\times{}^{12}C_5\left(\frac{1}{4}\right)^5\left(=\frac{99}{16}\right)\)
B1ft
Must be independent of \(x\); most candidates do not find the additional term so typically score max 1/3 in (b)
M1 for attempting to add \("\frac{55}{16}"\) to \(8\times{}^{12}C_5\left(\frac{1}{4}\right)^5\); A1 for \(\frac{77}{8}\) oe e.g. 9.625
## Question 3:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(1+\frac{x}{4}\right)^{12} = 1+12\left(\frac{x}{4}\right)^1 + \frac{12\times11}{2}\left(\frac{x}{4}\right)^2 + \frac{12\times11\times10}{3!}\left(\frac{x}{4}\right)^3+...$ | M1 | Correct attempt at binomial expansion; accept sight of $^{12}C_2\left(\frac{x}{4}\right)^2$ or $^{12}C_3\left(\frac{x}{4}\right)^3$ condoning omission of brackets. FYI $^{12}C_2=66$, $^{12}C_3=220$ |
| $= 1+3x+\frac{33}{8}x^2+\frac{55}{16}x^3$ | B1, A1 | B1 for $1+3x$ must be simplified; A1 for $+\frac{33}{8}x^2+\frac{55}{16}x^3$, accept $+4.125x^2+3.4375x^3$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sight of term independent of $x$: $"\frac{55}{16}"$ or $8\times{}^{12}C_5\left(\frac{1}{4}\right)^5\left(=\frac{99}{16}\right)$ | B1ft | Must be independent of $x$; most candidates do not find the additional term so typically score max 1/3 in (b) |
| $"\frac{55}{16}"+8\times{}^{12}C_5\left(\frac{1}{4}\right)^5 = \frac{55}{16}+\frac{99}{16}=\frac{77}{8}$ | M1, A1 | M1 for attempting to add $"\frac{55}{16}"$ to $8\times{}^{12}C_5\left(\frac{1}{4}\right)^5$; A1 for $\frac{77}{8}$ oe e.g. 9.625 |
---
3. (a) Find the first 4 terms, in ascending powers of $x$, in the binomial expansion of
$$\left( 1 + \frac { x } { 4 } \right) ^ { 12 }$$
giving each coefficient in its simplest form.\\
(b) Find the term independent of $x$ in the expansion of
$$\left( \frac { x ^ { 2 } + 8 } { x ^ { 5 } } \right) \left( 1 + \frac { x } { 4 } \right) ^ { 12 }$$
\hfill \mbox{\textit{Edexcel P2 2019 Q3 [6]}}