WJEC Further Unit 1 Specimen — Question 3 6 marks

Exam BoardWJEC
ModuleFurther Unit 1 (Further Unit 1)
SessionSpecimen
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeStandard summation formulae application
DifficultyChallenging +1.2 This is a standard Further Maths summation problem requiring the method of differences or direct summation of a cubic polynomial. While it involves algebraic manipulation across multiple steps (expanding the general term, applying summation formulas for powers, and factorizing), the technique is well-practiced in Further Maths courses. The 6-mark allocation and straightforward structure place it above average difficulty but well within expected Further Maths competency.
Spec4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series

Find an expression, in terms of \(n\), for the sum of the first \(n\) terms of the series $$1.2.4 + 2.3.5 + 3.4.6 + \ldots + n(n + 1)(n + 3) + \ldots$$ Express your answer as a product of linear factors. [6]

AnswerMarks Guidance
\[S_n = \sum_{r=1}^{n} r(r+1)(r+3)\]M1 AO1
\[= \sum_{r=1}^{n} (r^3 + 4r^2 + 3r)\]A1 AO1
\[= \frac{n^2(n+1)^2}{4} + 4\frac{n(n+1)(2n+1)}{6} + 3\frac{n(n+1)}{2}\]A1 AO1
\[= \frac{n(n+1)}{12}(3n(n+1) + 8(2n+1) + 18)\]A1 AO1
\[= \frac{n(n+1)}{12}(3n^2 + 19n + 26)\]A1 AO1
\[= \frac{n(n+1)(n+2)(3n+13)}{12}\]A1 AO1
Total: [6]
Question 4(a)
AnswerMarks Guidance
\[\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\beta\gamma + \gamma\alpha + \alpha\beta)\]M1 AO2
\[= 4^2 - 2 \times 14 = -12\]A1 AO2
A cubic equation either has 3 real roots or 1 real root. Since the sum of squares is negative, the 3 roots cannot all be real so there is just 1 real root.B1 AO2
Question 4(b)
AnswerMarks Guidance
A second root is \(1 - 3i\), since complex roots occur in conjugate pairs.B1 AO2
The third root is 2 since the sum of the 2 complex roots is 2 and the sum of the 3 roots is 4.E1, B1 AO2, AO1
Total: [7]
$$S_n = \sum_{r=1}^{n} r(r+1)(r+3)$$ | M1 | AO1

$$= \sum_{r=1}^{n} (r^3 + 4r^2 + 3r)$$ | A1 | AO1

$$= \frac{n^2(n+1)^2}{4} + 4\frac{n(n+1)(2n+1)}{6} + 3\frac{n(n+1)}{2}$$ | A1 | AO1

$$= \frac{n(n+1)}{12}(3n(n+1) + 8(2n+1) + 18)$$ | A1 | AO1

$$= \frac{n(n+1)}{12}(3n^2 + 19n + 26)$$ | A1 | AO1

$$= \frac{n(n+1)(n+2)(3n+13)}{12}$$ | A1 | AO1

| **Total: [6]** |

# Question 4(a)

$$\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\beta\gamma + \gamma\alpha + \alpha\beta)$$ | M1 | AO2

$$= 4^2 - 2 \times 14 = -12$$ | A1 | AO2

A cubic equation either has 3 real roots or 1 real root. Since the sum of squares is negative, the 3 roots cannot all be real so there is just 1 real root. | B1 | AO2

# Question 4(b)

A second root is $1 - 3i$, since complex roots occur in conjugate pairs. | B1 | AO2

The third root is 2 since the sum of the 2 complex roots is 2 and the sum of the 3 roots is 4. | E1, B1 | AO2, AO1

| **Total: [7]** |
Find an expression, in terms of $n$, for the sum of the first $n$ terms of the series
$$1.2.4 + 2.3.5 + 3.4.6 + \ldots + n(n + 1)(n + 3) + \ldots$$

Express your answer as a product of linear factors. [6]

\hfill \mbox{\textit{WJEC Further Unit 1  Q3 [6]}}