WJEC Further Unit 1 2023 June — Question 10 8 marks

Exam BoardWJEC
ModuleFurther Unit 1 (Further Unit 1)
Year2023
SessionJune
Marks8
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Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeAlternating series summation
DifficultyChallenging +1.2 This is a Further Maths question requiring pattern recognition and algebraic manipulation of an alternating series. Students must group consecutive pairs, recognize that each pair (2n-1)³ - (2n)³ simplifies using difference of cubes, sum the resulting arithmetic-like series, and express the answer in terms of k. While it requires multiple steps and some insight into pairing terms, the techniques are standard for Further Maths students and the question provides clear structure.
Spec4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series
10. Gareth is investigating a series involving cube numbers. His series is $$1 ^ { 3 } - 2 ^ { 3 } + 3 ^ { 3 } - 4 ^ { 3 } + 5 ^ { 3 } - 6 ^ { 3 } + 7 ^ { 3 } - \ldots$$ Gareth continues his series and ends with an odd number.
Find and simplify an expression for the sum of Gareth's series in terms of \(k\), where \(k\) is the number of odd numbers in his series.
METHOD 1: Realisation of difference of two series of cubes
AnswerMarks Guidance
\(\sum_{r=1}^{k}(2r - 1)^3 - \sum_{r=1}^{k-1}(2r)^3\)B1
\(= \sum_{r=1}^{k}(8r^3 - 12r^2 + 6r - 1) - \sum_{r=1}^{k-1}8r^3\)M1 Use of \(\Sigma\); Condone ranges for \(r\) other than \(r = k\) and \(r = k-1\) for ranges with a difference of 1
A1Cubing (ignore ranges)
\(= \frac{8}{4}k^2(k + 1)^2 - \frac{12}{6}k(k + 1)(2k + 1) + \frac{5}{2}k(k + 1) - k - \frac{4}{4}(k - 1)^2k^2\)m1 Use of sums formulae; All correct
A1
\(= k[2k^3 + 4k^2 + 2k - 4k^2 - 6k - 2 + 3k + 3 - 1 - 2k^3 + 4k^2 - 2k]\)A1 Simplification
\(= k^2(4k - 3)\)A1 Accept \(k(4k^2 - 3k)\) or \(4k^3 - 3k^2\)
METHOD 2:
\(1^3 - 2^3 + 3^3 - 4^3 + 5^3 - 6^3 + 7^3 - \ldots\)
AnswerMarks Guidance
\(1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + \ldots - 2(2^3 + 4^3 + 6^3 + \ldots)\)(B1) Realisation of difference of sequences
(B1)Factorising \(2^3\)
\(1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + \ldots (2k - 1 \text{ terms})\)
\(-16(1^3 + 2^3 + 3^3 + \ldots) (k - 1 \text{ terms})\)
AnswerMarks Guidance
\(\sum_{r=1}^{2k-1}r^3 - 16\sum_{r=1}^{k-1}r^3\)(M1) Use of \(\Sigma\); Condone ranges for \(r\) other than \(r = 2k-1\) and \(r = k-1\) for ranges with a difference of \(k\)
(A1)
\(= \frac{(2k - 1)^2(2k)^2}{4} - \frac{16(k - 1)^2k^2}{4}\)(m1) Use of sums formulae; All correct
(A1)
\(= k^2[(2k - 1)^2 - 4(k - 1)^2]\)(A1) Simplification
\(= k^2(4k^2 - 4k + 1 - 4k^2 + 8k - 4)\)
\(= k^2(4k - 3)\)(A1) Accept \(k(4k^2 - 3k)\) or \(4k^3 - 3k^2\)
METHOD 3: Realisation of difference of two series of cubes
AnswerMarks Guidance
\(\sum_{r=1}^{k}(2r - 1)^3 - \sum_{r=1}^{k-1}(2r)^3\)(B1)
\(= \sum_{r=1}^{k}(8r^3 - 12r^2 + 6r - 1) - \sum_{r=1}^{k-1}8r^3\)(A1) Cubing (ignore ranges)
\(= \sum_{r=k}^{k}8r^3 + \sum_{r=1}^{k}(-12r^2 + 6r - 1)\)(A1)
\(= 8k^3 - \frac{12}{6}k(k + 1)(2k + 1) + \frac{5}{2}k(k + 1) - k\)(m1) Use of sums formulae; All correct
(A1)
\(= k[8k^2 - 4k - 2 + 3k + 3 - 1]\)(A1) Simplification
\(= k^2(4k - 3)\)(A1) Accept \(k(4k^2 - 3k)\) or \(4k^3 - 3k^2\)
Total: [8]
**METHOD 1: Realisation of difference of two series of cubes**

$\sum_{r=1}^{k}(2r - 1)^3 - \sum_{r=1}^{k-1}(2r)^3$ | B1 |

$= \sum_{r=1}^{k}(8r^3 - 12r^2 + 6r - 1) - \sum_{r=1}^{k-1}8r^3$ | M1 | Use of $\Sigma$; Condone ranges for $r$ other than $r = k$ and $r = k-1$ for ranges with a difference of 1

| A1 | Cubing (ignore ranges)

$= \frac{8}{4}k^2(k + 1)^2 - \frac{12}{6}k(k + 1)(2k + 1) + \frac{5}{2}k(k + 1) - k - \frac{4}{4}(k - 1)^2k^2$ | m1 | Use of sums formulae; All correct

| A1 |

$= k[2k^3 + 4k^2 + 2k - 4k^2 - 6k - 2 + 3k + 3 - 1 - 2k^3 + 4k^2 - 2k]$ | A1 | Simplification

$= k^2(4k - 3)$ | A1 | Accept $k(4k^2 - 3k)$ or $4k^3 - 3k^2$

**METHOD 2:**

$1^3 - 2^3 + 3^3 - 4^3 + 5^3 - 6^3 + 7^3 - \ldots$

$1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + \ldots - 2(2^3 + 4^3 + 6^3 + \ldots)$ | (B1) | Realisation of difference of sequences

| (B1) | Factorising $2^3$

$1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + \ldots (2k - 1 \text{ terms})$ | | 

$-16(1^3 + 2^3 + 3^3 + \ldots) (k - 1 \text{ terms})$

$\sum_{r=1}^{2k-1}r^3 - 16\sum_{r=1}^{k-1}r^3$ | (M1) | Use of $\Sigma$; Condone ranges for $r$ other than $r = 2k-1$ and $r = k-1$ for ranges with a difference of $k$

| (A1) |

$= \frac{(2k - 1)^2(2k)^2}{4} - \frac{16(k - 1)^2k^2}{4}$ | (m1) | Use of sums formulae; All correct

| (A1) |

$= k^2[(2k - 1)^2 - 4(k - 1)^2]$ | (A1) | Simplification

$= k^2(4k^2 - 4k + 1 - 4k^2 + 8k - 4)$ | | 

$= k^2(4k - 3)$ | (A1) | Accept $k(4k^2 - 3k)$ or $4k^3 - 3k^2$

**METHOD 3: Realisation of difference of two series of cubes**

$\sum_{r=1}^{k}(2r - 1)^3 - \sum_{r=1}^{k-1}(2r)^3$ | (B1) |

$= \sum_{r=1}^{k}(8r^3 - 12r^2 + 6r - 1) - \sum_{r=1}^{k-1}8r^3$ | (A1) | Cubing (ignore ranges)

$= \sum_{r=k}^{k}8r^3 + \sum_{r=1}^{k}(-12r^2 + 6r - 1)$ | (A1) |

$= 8k^3 - \frac{12}{6}k(k + 1)(2k + 1) + \frac{5}{2}k(k + 1) - k$ | (m1) | Use of sums formulae; All correct

| (A1) |

$= k[8k^2 - 4k - 2 + 3k + 3 - 1]$ | (A1) | Simplification

$= k^2(4k - 3)$ | (A1) | Accept $k(4k^2 - 3k)$ or $4k^3 - 3k^2$

**Total: [8]**
10. Gareth is investigating a series involving cube numbers. His series is

$$1 ^ { 3 } - 2 ^ { 3 } + 3 ^ { 3 } - 4 ^ { 3 } + 5 ^ { 3 } - 6 ^ { 3 } + 7 ^ { 3 } - \ldots$$

Gareth continues his series and ends with an odd number.\\
Find and simplify an expression for the sum of Gareth's series in terms of $k$, where $k$ is the number of odd numbers in his series.

\hfill \mbox{\textit{WJEC Further Unit 1 2023 Q10 [8]}}
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