OCR H240/01 2022 June — Question 6 8 marks

Exam BoardOCR
ModuleH240/01 (Pure Mathematics)
Year2022
SessionJune
Marks8
PaperDownload PDF ↗
TopicBinomial Theorem (positive integer n)
TypeSubstitution into binomial expansion
DifficultyStandard +0.3 Part (a) is a straightforward binomial expansion with small integer n=5, requiring only routine application of the formula. Part (b) adds a substitution step (letting x = y + (4/3)y² or similar) and coefficient matching, which requires some algebraic manipulation but follows a standard pattern once the connection is recognized. This is slightly above average difficulty due to the two-part structure and the need to connect the parts, but remains a textbook-style question.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n

6
  1. Find the first four terms in the expansion of \(( 3 + 2 x ) ^ { 5 }\) in ascending powers of \(x\).
  2. Hence determine the coefficient of \(y ^ { 3 }\) in the expansion of \(\left( 3 + 2 y + 4 y ^ { 2 } \right) ^ { 5 }\).

Question 6(a):
AnswerMarks Guidance
AnswerMark Guidance
\(3^5 + 5\times3^4\times(2x) = 243 + 810x\)B1 Obtain \(243 + 810x\); condone \(3^5 + 810x\); allow terms not written as a sum e.g. written separately or linked with a comma
\(10\times3^3\times(2x)^2\) or \(10\times3^2\times(2x)^3\)M1 Attempt at least one further term — product of correct binomial coeff, power of 3 and attempted power of \(2x\), with powers totalling 5; binomial coeff must be numerical; \(^5C_2\) is not yet enough; allow BOD if brackets missing when index is applied to \(2x\), even if never recovered e.g. \(540x^2\) or \(180x^3\)
\(+ 1080x^2\)A1 Obtain correct third term; coefficient simplified; terms separate, listed or summed
\(+ 720x^3\)A1 [4] Obtain correct fourth term; coefficient simplified; could be separate term, part of a list or part of a sum; if expanding brackets then mark as above, but all 5 sets of brackets must be considered (allow irrelevant terms to be discarded)
Alternative method — Expanding \(\left[3\left(1+\frac{2}{3}x\right)\right]^5\):
AnswerMarks Guidance
AnswerMark Guidance
\(243 + 810x\) or \(243\left(1+\frac{10}{3}x\right)\)B1 First two terms correct; allow with 243 still outside the bracket
\(243\left(\frac{40}{9}x^2\right)\) or \(243\left(\frac{80}{27}x^3\right)\)M1 Attempt one further term; condone just 3 not \(3^5\) being used, but must be the correct binomial coeff and an attempt at the correct power of \(\frac{2}{3}x\); allow BOD if no brackets
\(243 + 810x + 1080x^2 + 720x^3\)A1, A1 [4] Either \(3^{rd}\) or \(4^{th}\) term correct; fully correct expansion; with the 243 now multiplied into the expansion
Question 6(b):
AnswerMarks Guidance
AnswerMark Guidance
\(x = y + 2y^2\)B1 Identify correct substitution; could be stated or implied by use in their binomial expansion
\(1080(y+2y^2)^2 + 720(y+2y^2)^3\)M1 Attempt to use binomial from (a) with their 2 term substitution; must substitute into at least the \(x^2\) and \(x^3\) terms from their (a); allow M1 if using \(2y + 4y^2\) as their substitution
\(4320y^3 + 720y^3\)M1 Attempt expansion to obtain the two relevant terms in \(y^3\); M0 if any other \(y^3\) terms; expect 4(their 1080) and (their 720); allow M1 if using \(2y+4y^2\) as substitution — expect 16(their 1080) and 8(their 720)
coeff of \(y^3\) is 5040A1 [4] Allow \(5040y^3\); ignore any other non-cubic terms
Alternative method 1 — Attempting binomial expansion of \((3+(2y+4y^2))^5\) or \(((3+2y)+4y^2)^5\):
AnswerMarks Guidance
AnswerMark Guidance
e.g. \(((3+2y)+(4y^2))\)B1 Group into two expressions and attempt to use them
e.g. \((3+2y)^5 + 5(3+2y)^4(4y^2)\)M1 Use their groups to obtain the appropriate two elements of their binomial expansion (i.e. those that would give \(y^3\) terms)
e.g. \((\ldots + 720y^3 + \ldots) + 5(\ldots 4\cdot3^3\cdot2y\ldots)(4y^2) = 720y^3 + 4320y^3\)M1 Expand to attempt the two \(y^3\) terms and no others
coeff of \(y^3\) is 5040A1 Obtain 5040
Alternative method 2 — Attempting to expand all 5 brackets:
AnswerMarks Guidance
AnswerMark Guidance
e.g. \((3+2y+4y^2)^5 = (81+216y+648y^2+960y^3\ldots)(3+2y+4y^2)\)M1 Attempt to use all 5 brackets; an attempt to use all 5 is sufficient
\((216y\times4y^2)+(648y^2\times2y)+(960y^3\times3)\)M1 Attempt all products that would give a \(y\)-cubed term; condone additional terms even those that would give another \(y^3\) term; irrelevant terms (i.e. powers greater than 3) may never be seen
Question 6 (Binomial expansion):
AnswerMarks Guidance
AnswerMarks Guidance
\(864y^3 + 1296y^3 + 2880y^3\)A1 Obtain correct terms or coefficients, with no more than one incorrect. Must have attempted all expected \(y^3\) terms, no more than one coefficient error. If \((3+2y+4y^2)^4 \times (3+2y+4y^2)\) then expect \(2880+1296+864\). If \((3+2y+4y^2)^3 \times (3+2y+4y^2)^2\) then expect \(1368+1728+1512+432\)
coeff of \(y^3\) is \(5040\)A1 Obtain \(5040\)
[4]
# Question 6(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| $3^5 + 5\times3^4\times(2x) = 243 + 810x$ | B1 | Obtain $243 + 810x$; condone $3^5 + 810x$; allow terms not written as a sum e.g. written separately or linked with a comma |
| $10\times3^3\times(2x)^2$ or $10\times3^2\times(2x)^3$ | M1 | Attempt at least one further term — product of correct binomial coeff, power of 3 and attempted power of $2x$, with powers totalling 5; binomial coeff must be numerical; $^5C_2$ is not yet enough; allow BOD if brackets missing when index is applied to $2x$, even if never recovered e.g. $540x^2$ or $180x^3$ |
| $+ 1080x^2$ | A1 | Obtain correct third term; coefficient simplified; terms separate, listed or summed |
| $+ 720x^3$ | A1 [4] | Obtain correct fourth term; coefficient simplified; could be separate term, part of a list or part of a sum; if expanding brackets then mark as above, but all 5 sets of brackets must be considered (allow irrelevant terms to be discarded) |

**Alternative method** — Expanding $\left[3\left(1+\frac{2}{3}x\right)\right]^5$:

| Answer | Mark | Guidance |
|--------|------|----------|
| $243 + 810x$ or $243\left(1+\frac{10}{3}x\right)$ | B1 | First two terms correct; allow with 243 still outside the bracket |
| $243\left(\frac{40}{9}x^2\right)$ or $243\left(\frac{80}{27}x^3\right)$ | M1 | Attempt one further term; condone just 3 not $3^5$ being used, but must be the correct binomial coeff and an attempt at the correct power of $\frac{2}{3}x$; allow BOD if no brackets |
| $243 + 810x + 1080x^2 + 720x^3$ | A1, A1 [4] | Either $3^{rd}$ or $4^{th}$ term correct; fully correct expansion; with the 243 now multiplied into the expansion |

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# Question 6(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $x = y + 2y^2$ | B1 | Identify correct substitution; could be stated or implied by use in their binomial expansion |
| $1080(y+2y^2)^2 + 720(y+2y^2)^3$ | M1 | Attempt to use binomial from (a) with their 2 term substitution; must substitute into at least the $x^2$ and $x^3$ terms from their (a); allow **M1** if using $2y + 4y^2$ as their substitution |
| $4320y^3 + 720y^3$ | M1 | Attempt expansion to obtain the two relevant terms in $y^3$; **M0** if any other $y^3$ terms; expect 4(their 1080) and (their 720); allow **M1** if using $2y+4y^2$ as substitution — expect 16(their 1080) and 8(their 720) |
| coeff of $y^3$ is 5040 | A1 [4] | Allow $5040y^3$; ignore any other non-cubic terms |

**Alternative method 1** — Attempting binomial expansion of $(3+(2y+4y^2))^5$ or $((3+2y)+4y^2)^5$:

| Answer | Mark | Guidance |
|--------|------|----------|
| e.g. $((3+2y)+(4y^2))$ | B1 | Group into two expressions and attempt to use them |
| e.g. $(3+2y)^5 + 5(3+2y)^4(4y^2)$ | M1 | Use their groups to obtain the appropriate two elements of their binomial expansion (i.e. those that would give $y^3$ terms) |
| e.g. $(\ldots + 720y^3 + \ldots) + 5(\ldots 4\cdot3^3\cdot2y\ldots)(4y^2) = 720y^3 + 4320y^3$ | M1 | Expand to attempt the two $y^3$ terms and no others |
| coeff of $y^3$ is 5040 | A1 | Obtain 5040 |

**Alternative method 2** — Attempting to expand all 5 brackets:

| Answer | Mark | Guidance |
|--------|------|----------|
| e.g. $(3+2y+4y^2)^5 = (81+216y+648y^2+960y^3\ldots)(3+2y+4y^2)$ | M1 | Attempt to use all 5 brackets; an attempt to use all 5 is sufficient |
| $(216y\times4y^2)+(648y^2\times2y)+(960y^3\times3)$ | M1 | Attempt all products that would give a $y$-cubed term; condone additional terms even those that would give another $y^3$ term; irrelevant terms (i.e. powers greater than 3) may never be seen |

# Question 6 (Binomial expansion):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $864y^3 + 1296y^3 + 2880y^3$ | A1 | Obtain correct terms or coefficients, with no more than one incorrect. Must have attempted all expected $y^3$ terms, no more than one coefficient error. If $(3+2y+4y^2)^4 \times (3+2y+4y^2)$ then expect $2880+1296+864$. If $(3+2y+4y^2)^3 \times (3+2y+4y^2)^2$ then expect $1368+1728+1512+432$ |
| coeff of $y^3$ is $5040$ | A1 | Obtain $5040$ |
| | **[4]** | |

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6
\begin{enumerate}[label=(\alph*)]
\item Find the first four terms in the expansion of $( 3 + 2 x ) ^ { 5 }$ in ascending powers of $x$.
\item Hence determine the coefficient of $y ^ { 3 }$ in the expansion of $\left( 3 + 2 y + 4 y ^ { 2 } \right) ^ { 5 }$.
\end{enumerate}

\hfill \mbox{\textit{OCR H240/01 2022 Q6 [8]}}