Moderate -0.8 This is a straightforward C2 binomial expansion question requiring routine application of the binomial theorem formula and simple algebraic manipulation. Part (a) is direct substitution into the formula, and part (b) involves setting up and solving a simple linear equation. The question requires no problem-solving insight beyond standard textbook methods.
2. (a) Find the first 3 terms, in ascending powers of \(x\), of the binomial expansion of
$$( 2 + k x ) ^ { 7 }$$
where \(k\) is a constant. Give each term in its simplest form.
Given that the coefficient of \(x ^ { 2 }\) is 6 times the coefficient of \(x\),
(b) find the value of \(k\).
(If \(672kx^2\) follows \(672(kx)^2\), isw and allow A1)
Part (b)
Answer
Marks
Guidance
\(6 \times 448k = 672k^2\)
M1
\(k = 4\)
A1
(Ignore \(k = 0\), if seen)
Guidance:
M1 for either the \(x\) term or the \(x^2\) term. Requires correct binomial coefficient in any form with the correct power of the coefficient (perhaps including powers of 2 and/or \(k\)) may be wrong or missing.
Allow binomial coefficients such as \(\binom{7}{1}\), \(\binom{7}{1}\), \(\binom{7}{2}\), \(C_1'\), \(C_2'\).
However, \(448 + kx\) or similar is M0.
B1, A1, A1 for the simplified versions seen above.
Alternative: Note that a factor \(2^7\) can be taken out first: \(2^7\left(1 + \frac{kx}{2}\right)^7\), but the mark scheme still applies.
Answer
Marks
Guidance
Ignoring subsequent working (isw): isw if necessary after correct working: e.g. \(128 + 448kx + 672k^2x^2\)
M1 B1 A1 A1 = \(4 + 14kx + 21k^2x^2\)
isw (Full marks are still available in part (b)).
M1 for equating their coefficient of \(x^2\) to 6 times that of \(x... \) to get an equation in \(k\), ... or equating their coefficient of \(x\) to 6 times that of \(x^2\), to get an equation in \(k\).
Allow this M mark even if the equation is trivial, providing their coefficients from part (a) have been used, e.g. \(6 \times 448k = 672k\), but beware \(k = 4\) following from this, which is A0.
An equation in \(k\) alone is required for this M mark, so... e.g. \(6 \times 448kx = 672k^2x^2 \Rightarrow k = 4\) or similar is M0 A0 (equation in coefficients only is never seen), but... e.g. \(6 \times 448kx = 672k^2x^2 \Rightarrow 6 \times 448k = 672k^2 \Rightarrow k = 4\) will get M1 A1 (as coefficients rather than terms have now been considered).
The mistake \(2\left(1 + \frac{kx}{2}\right)^7\) would give a maximum of 3 marks: M1B0A0, M1A1.
**Part (a)**
$(7 \times ... \times x)$ or $(21 \times ... \times x^2)$ | M1 | The 7 or 21 can be in 'unsimplified' form.
$(2 + kx)^7 = 2^7 + 2^6 \times 7 \times kx + 2^5 \times \binom{7}{2} k^2 x^2 = 128 + 448kx + 672k^2x^2$ [or $672(kx)^2$] | B1, A1, A1 | (If $672kx^2$ follows $672(kx)^2$, isw and allow A1)
**Part (b)**
$6 \times 448k = 672k^2$ | M1 |
$k = 4$ | A1 | (Ignore $k = 0$, if seen)
**Guidance:**
M1 for either the $x$ term or the $x^2$ term. Requires correct binomial coefficient in any form with the correct power of the coefficient (perhaps including powers of 2 and/or $k$) may be wrong or missing.
Allow binomial coefficients such as $\binom{7}{1}$, $\binom{7}{1}$, $\binom{7}{2}$, $C_1'$, $C_2'$.
However, $448 + kx$ or similar is M0.
B1, A1, A1 for the simplified versions seen above.
**Alternative:** Note that a factor $2^7$ can be taken out first: $2^7\left(1 + \frac{kx}{2}\right)^7$, but the mark scheme still applies.
Ignoring subsequent working (isw): isw if necessary after correct working: e.g. $128 + 448kx + 672k^2x^2$ | M1 B1 A1 A1 = $4 + 14kx + 21k^2x^2$ | isw (Full marks are still available in part (b)).
M1 for equating their coefficient of $x^2$ to 6 times that of $x... $ to get an equation in $k$, ... or equating their coefficient of $x$ to 6 times that of $x^2$, to get an equation in $k$.
Allow this M mark even if the equation is trivial, providing their coefficients from part (a) have been used, e.g. $6 \times 448k = 672k$, but beware $k = 4$ following from this, which is A0.
An equation in $k$ alone is required for this M mark, so... e.g. $6 \times 448kx = 672k^2x^2 \Rightarrow k = 4$ or similar is M0 A0 (equation in coefficients only is never seen), but... e.g. $6 \times 448kx = 672k^2x^2 \Rightarrow 6 \times 448k = 672k^2 \Rightarrow k = 4$ will get M1 A1 (as coefficients rather than terms have now been considered).
The mistake $2\left(1 + \frac{kx}{2}\right)^7$ would give a maximum of 3 marks: M1B0A0, M1A1.
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2. (a) Find the first 3 terms, in ascending powers of $x$, of the binomial expansion of
$$( 2 + k x ) ^ { 7 }$$
where $k$ is a constant. Give each term in its simplest form.
Given that the coefficient of $x ^ { 2 }$ is 6 times the coefficient of $x$,\\
(b) find the value of $k$.\\
\hfill \mbox{\textit{Edexcel C2 2009 Q2 [6]}}