Moderate -0.8 This is a straightforward application of the binomial theorem requiring routine expansion and algebraic simplification. Part (a) is direct substitution into the binomial formula, and part (b) involves recognizing that odd powers cancel when adding the two expansions. No problem-solving insight is needed, making this easier than average.
4. (a) Find the first four terms, in ascending powers of \(x\), of the binomial expansion of
$$\left( 2 - \frac { 1 } { 4 } x \right) ^ { 6 }$$
(b) Given that \(x\) is small, so terms in \(x ^ { 4 }\) and higher powers of \(x\) may be ignored, show
$$\left( 2 - \frac { 1 } { 4 } x \right) ^ { 6 } + \left( 2 + \frac { 1 } { 4 } x \right) ^ { 6 } = a + b x ^ { 2 }$$
where \(a\) and \(b\) are constants to be found.
For either \(2^6\) or 64. Award for an unsimplified \({^6C_0 2^6} \left(-\frac{1}{4}x\right)^0\). M1: For an attempt at the binomial expansion. Score for a correct attempt at term 2, 3 or 4. Accept sight of \({^6C_1 2^5} \left(\pm\frac{1}{4}x\right)\), \({^6C_2 2^4} \left(\pm\frac{1}{4}x\right)^2\), \({^6C_3}\left(\pm\frac{1}{4}x\right)^3\) condoning omission of brackets.
\(= 64 - 48x + 15x^2 - 2.5x^3\)
A1 A1
For any two simplified terms of \(-48x + 15x^2 - 2.5x^3\). For \(64 - 48x + 15x^2 - 2.5x^3\) ignoring terms with greater powers. This may be awarded in (b) if it is not fully simplified in (a). Allow the terms to be listed \(64, -48x, 15x^2, -2.5x^3\). Isw after sight of correct values. The expression written out without any method can be awarded all 4 marks.
For adding two sequences that must be of the correct form with the correct signs. Look for \(\left(A - Bx + Cx^2 - Dx^3\right) + \left(A + Bx + Cx^2 + Dx^3\right)\) but condone \(\left(A - Bx + Cx^2\right) + \left(A + Bx + Cx^2\right)\). For this to be scored there must be some negative terms in (a).
\(\approx 128 + 30x^2\)
B1ft A1
For one correct term (follow through). Usually \(a = 128\) but accept either \(a = 2 \times\) 'their' +ve 64 or \(b = 2 \times\) 'their' + ve 15. CSO so must be from \((64 - 48x + 15x^2 - 2.5x^3) + (64 + 48x + 15x^2 + 2.5x^3)\). Allow \(a = 128, b = 30\) following correct work. This is a show that question so M1 must be awarded. It must be their final answer so do not isw here.
(3 marks, 7 marks total)
Additional notes for (b):
M1: For adding two sequences that must be of the correct form with the correct signs. Look for \(\left(A - Bx + Cx^2 - Dx^3\right) + \left(A + Bx + Cx^2 + Dx^3\right)\) but condone \(\left(A - Bx + Cx^2\right) + \left(A + Bx + Cx^2\right)\)
For this to be scored there must be some negative terms in (a)
B1ft: For one correct term (follow through). Usually \(a = 128\) but accept either \(a = 2 \times\) 'their' +ve 64 or \(b = 2 \times\) 'their' + ve 15
A1: For \(128 + 30x^2\). CSO so must be from \(\left(64 - 48x + 15x^2 - 2.5x^3\right) + \left(64 + 48x + 15x^2 + 2.5x^3\right)\). Allow \(a = 128, b = 30\) following correct work. This is a show that question so M1 must be awarded. It must be their final answer so do not isw here.
M1: For an attempt at the binomial expansion seen in at least one term within the brackets. Score for a correct attempt at term 2, 3 or 4.
Accept sight of \(6\left(\pm\frac{1}{8}x\right)\), \(\frac{6 \times 5}{2}\left(\pm\frac{1}{8}x\right)^2\), \(\frac{6 \times 5 \times 4}{3!}\left(\pm\frac{1}{8}x\right)^3\) condoning omission of brackets
A1: For any two terms of \(64 - 48x + 15x^2 - 2.5x^3\)
A1: For all four terms \(64 - 48x + 15x^2 - 2.5x^3\) ignoring terms with greater powers
Attempts to multiply out:
B1: For 64
M1: Multiplies out to form \(a + bx + cx^2 + dx^3 + ...\) and gets \(b, c\) or \(d\) correct.
A1A1: As main scheme
**(a)** $\left(2 - \frac{1}{4}x\right)^6 = 2^6 + {^6C_1 2^5 \left(-\frac{1}{4}x\right)} + {^6C_2 2^4} \left(-\frac{1}{4}x\right)^2 + {^6C_3 2^3} \left(-\frac{1}{4}x\right)^3 + {^6C_2 2^3} \left(-\frac{1}{4}x\right)^3 + ...$ | B1, M1 | For either $2^6$ or 64. Award for an unsimplified ${^6C_0 2^6} \left(-\frac{1}{4}x\right)^0$. M1: For an attempt at the binomial expansion. Score for a correct attempt at term 2, 3 or 4. Accept sight of ${^6C_1 2^5} \left(\pm\frac{1}{4}x\right)$, ${^6C_2 2^4} \left(\pm\frac{1}{4}x\right)^2$, ${^6C_3}\left(\pm\frac{1}{4}x\right)^3$ condoning omission of brackets.
$= 64 - 48x + 15x^2 - 2.5x^3$ | A1 A1 | For any two simplified terms of $-48x + 15x^2 - 2.5x^3$. For $64 - 48x + 15x^2 - 2.5x^3$ ignoring terms with greater powers. This may be awarded in (b) if it is not fully simplified in (a). Allow the terms to be listed $64, -48x, 15x^2, -2.5x^3$. Isw after sight of correct values. The expression written out without any method can be awarded all 4 marks.
| (4 marks)
**(b)** $\left(2 - \frac{1}{4}x\right)^6 + \left(2 + \frac{1}{4}x\right)^6 = (64 - 48x + 15x^2 - 2.5x^3) + (64 + 48x + 15x^2 + 2.5x^3)$ | M1 | For adding two sequences that must be of the correct form with the correct signs. Look for $\left(A - Bx + Cx^2 - Dx^3\right) + \left(A + Bx + Cx^2 + Dx^3\right)$ but condone $\left(A - Bx + Cx^2\right) + \left(A + Bx + Cx^2\right)$. For this to be scored there must be some negative terms in (a).
$\approx 128 + 30x^2$ | B1ft A1 | For one correct term (follow through). Usually $a = 128$ but accept either $a = 2 \times$ 'their' +ve 64 or $b = 2 \times$ 'their' + ve 15. CSO so must be from $(64 - 48x + 15x^2 - 2.5x^3) + (64 + 48x + 15x^2 + 2.5x^3)$. Allow $a = 128, b = 30$ following correct work. This is a show that question so M1 must be awarded. It must be their final answer so do not isw here.
| (3 marks, 7 marks total)
**Additional notes for (b):**
M1: For adding two sequences that must be of the correct form with the correct signs. Look for $\left(A - Bx + Cx^2 - Dx^3\right) + \left(A + Bx + Cx^2 + Dx^3\right)$ but condone $\left(A - Bx + Cx^2\right) + \left(A + Bx + Cx^2\right)$
For this to be scored there must be some negative terms in (a)
B1ft: For one correct term (follow through). Usually $a = 128$ but accept either $a = 2 \times$ 'their' +ve 64 or $b = 2 \times$ 'their' + ve 15
A1: For $128 + 30x^2$. CSO so must be from $\left(64 - 48x + 15x^2 - 2.5x^3\right) + \left(64 + 48x + 15x^2 + 2.5x^3\right)$. Allow $a = 128, b = 30$ following correct work. This is a show that question so M1 must be awarded. It must be their final answer so do not isw here.
Alternative method in (a):
$\left(2 - \frac{1}{4}x\right)^6 = 2^6 \left(1 - \frac{1}{8}x\right)^6 = 2^6 \left(1 + 6\left(-\frac{1}{8}x\right) + \frac{6 \times 5}{2}\left(-\frac{1}{8}x\right)^2 + \frac{6 \times 5 \times 4}{3!}\left(-\frac{1}{8}x\right)^3 + ...\right)$
B1: For sight of factor of either $2^6$ or 64.
M1: For an attempt at the binomial expansion seen in at least one term within the brackets. Score for a correct attempt at term 2, 3 or 4.
Accept sight of $6\left(\pm\frac{1}{8}x\right)$, $\frac{6 \times 5}{2}\left(\pm\frac{1}{8}x\right)^2$, $\frac{6 \times 5 \times 4}{3!}\left(\pm\frac{1}{8}x\right)^3$ condoning omission of brackets
A1: For any two terms of $64 - 48x + 15x^2 - 2.5x^3$
A1: For all four terms $64 - 48x + 15x^2 - 2.5x^3$ ignoring terms with greater powers
Attempts to multiply out:
B1: For 64
M1: Multiplies out to form $a + bx + cx^2 + dx^3 + ...$ and gets $b, c$ or $d$ correct.
A1A1: As main scheme
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4. (a) Find the first four terms, in ascending powers of $x$, of the binomial expansion of
$$\left( 2 - \frac { 1 } { 4 } x \right) ^ { 6 }$$
(b) Given that $x$ is small, so terms in $x ^ { 4 }$ and higher powers of $x$ may be ignored, show
$$\left( 2 - \frac { 1 } { 4 } x \right) ^ { 6 } + \left( 2 + \frac { 1 } { 4 } x \right) ^ { 6 } = a + b x ^ { 2 }$$
where $a$ and $b$ are constants to be found.\\
\hfill \mbox{\textit{Edexcel P2 2019 Q4 [7]}}