Edexcel C12 2016 January — Question 7 7 marks

Exam BoardEdexcel
ModuleC12 (Core Mathematics 1 & 2)
Year2016
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Theorem (positive integer n)
TypeFind constant given one specific term
DifficultyModerate -0.8 This is a straightforward binomial expansion question requiring recall of the formula and basic algebraic manipulation. Part (a) is routine application of (1+kx)^8 expansion, and part (b) involves solving a simple equation for k. No problem-solving insight needed, just standard technique.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n

7. (a) Find the first 4 terms, in ascending powers of \(x\), of the binomial expansion of \(( 1 + k x ) ^ { 8 }\), where \(k\) is a non-zero constant. Give each term in its simplest form. Given that the coefficient of \(x ^ { 3 }\) in this expansion is 1512
(b) find the value of \(k\).

Question 7:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\((1+kx)^8 = 1 + \binom{8}{1}(kx) + \binom{8}{2}(kx)^2 + \binom{8}{3}(kx)^3...\)M1 Attempt at binomial expansion to get third and/or fourth term; correct binomial coefficient combined with correct power of \(x\); accept \(^8C_2\) or \(^8C_3\) notation
\(= 1 + 8kx + 28k^2x^2 + 56k^3x^3 + ...\)B1, A1, A1 B1 for \(1+8kx\); A1 for \(28k^2x^2\) or \(28(kx)^2\); A1 for \(56k^3x^3\) or \(56(kx)^3\); ignore extra higher power terms
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Sets \(56k^3 = 1512\) and obtains \(k^3 = \frac{1512}{56}\)M1 A1 M1: sets coefficient of \(x^3 = 1512\), obtains \(k^n = ...\) where \(n\) is 1 or 3; A1: \(k^3 = \frac{1512}{56}\) or equivalent e.g. 27
So \(k = 3\)A1 cao (\(\pm3\) is A0); part (b) marked independently of part (a)
## Question 7:

### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(1+kx)^8 = 1 + \binom{8}{1}(kx) + \binom{8}{2}(kx)^2 + \binom{8}{3}(kx)^3...$ | M1 | Attempt at binomial expansion to get third and/or fourth term; correct binomial coefficient combined with correct power of $x$; accept $^8C_2$ or $^8C_3$ notation |
| $= 1 + 8kx + 28k^2x^2 + 56k^3x^3 + ...$ | B1, A1, A1 | B1 for $1+8kx$; A1 for $28k^2x^2$ or $28(kx)^2$; A1 for $56k^3x^3$ or $56(kx)^3$; ignore extra higher power terms |

### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $56k^3 = 1512$ and obtains $k^3 = \frac{1512}{56}$ | M1 A1 | M1: sets coefficient of $x^3 = 1512$, obtains $k^n = ...$ where $n$ is 1 or 3; A1: $k^3 = \frac{1512}{56}$ or equivalent e.g. 27 |
| So $k = 3$ | A1 | cao ($\pm3$ is A0); part (b) marked independently of part (a) |

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7. (a) Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of $( 1 + k x ) ^ { 8 }$, where $k$ is a non-zero constant. Give each term in its simplest form.

Given that the coefficient of $x ^ { 3 }$ in this expansion is 1512\\
(b) find the value of $k$.\\

\hfill \mbox{\textit{Edexcel C12 2016 Q7 [7]}}