OCR C2 2013 January — Question 4 7 marks

Exam BoardOCR
ModuleC2 (Core Mathematics 2)
Year2013
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Theorem (positive integer n)
TypeSubstitution into binomial expansion
DifficultyModerate -0.3 Part (i) is a straightforward application of the binomial theorem with small n=5, requiring only careful arithmetic. Part (ii) requires the insight to substitute (3y + y²) for x, then identify which terms contribute to y³, but this is a standard 'hence' follow-up that students are trained to recognize. The multi-step nature and coefficient extraction elevate it slightly above pure recall, but it remains a routine C2 question.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n1.04b Binomial probabilities: link to binomial expansion

4
  1. Find the binomial expansion of \(( 2 + x ) ^ { 5 }\), simplifying the terms.
  2. Hence find the coefficient of \(y ^ { 3 }\) in the expansion of \(\left( 2 + 3 y + y ^ { 2 } \right) ^ { 5 }\).

Question 4:
Part (i)
AnswerMarks Guidance
AnswerMarks Guidance
\((2+x)^5 = 32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5\)M1* Attempt expansion resulting in at least 5 terms – products of powers of 2 and \(x\). Each term must be an attempt at a product, including binomial coeffs if used. Allow M1 for no, or incorrect, binomial coeffs. Powers of 2 and \(x\) must be intended to sum to 5 within each term (allow slips if intention correct). Allow M1 for powers of \(\frac{1}{2}x\) from expanding \(k(1+\frac{1}{2}x)^5\), any \(k\) (allow if powers only applied to \(x\) and not \(\frac{1}{2}\))
M1d*Attempt to use correct binomial coefficients. At least 5 correct from 1, 5, 10, 10, 5, 1 – allow missing or incorrect (but not if raised to a power). May be implied rather than explicit. Must be numerical e.g. \(^5C_1\) is not enough. They must be part of a product within each term. The coefficient must be used in an attempt at the relevant term ie \(5 \times 2^3 \times x^3\) is M0. Allow M1 for correct coeffs from \(k(1+\frac{1}{2}x)^5\), any \(k\)
A1Obtain at least 4 correct simplified terms. Either linked by '\(+\)' or as part of a list
A1Obtain a fully correct expansion. Terms must be linked by '\(+\)' and not just commas. A0 if a correct expansion is subsequently spoiled by attempt to simplify, including division. SR for expanding brackets: M2 for attempt using all 5 brackets giving a quintic; A1 obtain at least 4 correct simplified terms; A1 obtain a fully correct expansion
[4]
Part (ii)
AnswerMarks Guidance
AnswerMarks Guidance
\(80(3y+y^2)^2 + 40(3y+y^2)^3\), coeff of \(y^3 = (80 \times 6)+(40 \times 27) = 1560\)M1 Attempt to use \(x = 3y+y^2\). Replace \(x\) with \(3y+y^2\) in at least one relevant term and attempt expansion, including relevant numerical coeff from (i) or from restart
A1Obtain \(480(y^3)\) or \(1080(y^3)\). Could be with other terms, inc \(y^3\)
A1Obtain 1560 (or \(1560y^3\)). Ignore terms involving powers other than \(y^3\)
OR \((1+y)^5(2+y)^5 = (1+5y+10y^2+10y^3...)\times(32+80y+80y^2+40y^3...)\), coeff of \(y^3 = 320+800+400+40\)M1 Attempt expansion of both \((1+y)^5\) and \((2+y)^5\) (allow powers higher than 3 to be discarded) and make some attempt at the product
A1Obtain at least 2 correct coeffs of \(y^3\)
A1Obtain 1560 (or \(1560y^3\))
OR \(((2+3y)+y^2)^5 = (2+3y)^5 + 5(2+3y)^4y^2 = ...10 \times 4 \times 27y^3... + 5 \times 4 \times 8 \times 3y \times y^2\), coeff of \(y^3 = 1080+480 = 1560\)M1 Attempt expansion of at least one relevant term
A1Obtain \(480(y^3)\) or \(1080(y^3)\)
A1Obtain 1560 (or \(1560y^3\)). OR M1 attempt expansion of all 5 brackets (allow powers higher than 3 to be discarded throughout method); A2 obtain 1560 (or \(1560y^3\))
[3]
# Question 4:

## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2+x)^5 = 32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5$ | M1* | Attempt expansion resulting in at least 5 terms – products of powers of 2 and $x$. Each term must be an attempt at a product, including binomial coeffs if used. Allow M1 for no, or incorrect, binomial coeffs. Powers of 2 and $x$ must be intended to sum to 5 within each term (allow slips if intention correct). Allow M1 for powers of $\frac{1}{2}x$ from expanding $k(1+\frac{1}{2}x)^5$, any $k$ (allow if powers only applied to $x$ and not $\frac{1}{2}$) |
| | M1d* | Attempt to use correct binomial coefficients. At least 5 correct from 1, 5, 10, 10, 5, 1 – allow missing or incorrect (but not if raised to a power). May be implied rather than explicit. Must be numerical e.g. $^5C_1$ is not enough. They must be part of a product within each term. The coefficient must be used in an attempt at the relevant term ie $5 \times 2^3 \times x^3$ is M0. Allow M1 for correct coeffs from $k(1+\frac{1}{2}x)^5$, any $k$ |
| | A1 | Obtain at least 4 correct simplified terms. Either linked by '$+$' or as part of a list |
| | A1 | Obtain a fully correct expansion. Terms must be linked by '$+$' and not just commas. A0 if a correct expansion is subsequently spoiled by attempt to simplify, including division. **SR for expanding brackets:** M2 for attempt using all 5 brackets giving a quintic; A1 obtain at least 4 correct simplified terms; A1 obtain a fully correct expansion |
| | [4] | |

## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $80(3y+y^2)^2 + 40(3y+y^2)^3$, coeff of $y^3 = (80 \times 6)+(40 \times 27) = 1560$ | M1 | Attempt to use $x = 3y+y^2$. Replace $x$ with $3y+y^2$ in at least one relevant term and attempt expansion, including relevant numerical coeff from (i) or from restart |
| | A1 | Obtain $480(y^3)$ or $1080(y^3)$. Could be with other terms, inc $y^3$ |
| | A1 | Obtain 1560 (or $1560y^3$). Ignore terms involving powers other than $y^3$ |
| **OR** $(1+y)^5(2+y)^5 = (1+5y+10y^2+10y^3...)\times(32+80y+80y^2+40y^3...)$, coeff of $y^3 = 320+800+400+40$ | M1 | Attempt expansion of both $(1+y)^5$ and $(2+y)^5$ (allow powers higher than 3 to be discarded) and make some attempt at the product |
| | A1 | Obtain at least 2 correct coeffs of $y^3$ |
| | A1 | Obtain 1560 (or $1560y^3$) |
| **OR** $((2+3y)+y^2)^5 = (2+3y)^5 + 5(2+3y)^4y^2 = ...10 \times 4 \times 27y^3... + 5 \times 4 \times 8 \times 3y \times y^2$, coeff of $y^3 = 1080+480 = 1560$ | M1 | Attempt expansion of at least one relevant term |
| | A1 | Obtain $480(y^3)$ or $1080(y^3)$ |
| | A1 | Obtain 1560 (or $1560y^3$). **OR** M1 attempt expansion of all 5 brackets (allow powers higher than 3 to be discarded throughout method); A2 obtain 1560 (or $1560y^3$) |
| | [3] | |
4 (i) Find the binomial expansion of $( 2 + x ) ^ { 5 }$, simplifying the terms.\\
(ii) Hence find the coefficient of $y ^ { 3 }$ in the expansion of $\left( 2 + 3 y + y ^ { 2 } \right) ^ { 5 }$.

\hfill \mbox{\textit{OCR C2 2013 Q4 [7]}}