WJEC Further Unit 1 2022 June — Question 4 7 marks

Exam BoardWJEC
ModuleFurther Unit 1 (Further Unit 1)
Year2022
SessionJune
Marks7
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Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeFinding n for given sum value
DifficultyChallenging +1.2 This question requires knowing the sum of squares formula (n(n+1)(2n+1)/6), setting it equal to (3N-2)², then solving the resulting cubic equation. While it involves multiple steps and algebraic manipulation, the approach is straightforward once the formula is recalled, and the cubic likely factors nicely. It's moderately harder than average due to the cubic solving, but remains a standard Further Maths exercise without requiring novel insight.
Spec4.06a Summation formulae: sum of r, r^2, r^3
4. The positive integer \(N\) is such that \(1 ^ { 2 } + 2 ^ { 2 } + 3 ^ { 2 } + \ldots + N ^ { 2 } = ( 3 N - 2 ) ^ { 2 }\). Write down and simplify an equation satisfied by \(N\). Hence find the possible values of \(N\).
Question 4:
AnswerMarks Guidance
WorkingMark Guidance
\(\sum_{r=1}^{N} r^2 = (3N-2)^2\)
\(\frac{1}{6}N(N+1)(2N+1) = 9N^2-12N+4\)M1
\(2N^3+3N^2+N = 54N^2-72N+24\)A1
\(2N^3-51N^2+73N-24 = 0\)A1 cao
Finding one factor, e.g. \((N-1)\): \((N-1)(2N^2-49N+24) = 0\)B1, m1 \((N-k)\) form; Linear \(\times\) Quadratic (2 terms correct)
\(\therefore (N-1)(2N-1)(N-24) = 0\)
\(\therefore N=1\) or \(N=\frac{1}{2}\) or \(N=24\)A1
Therefore \(N = 1, 24\)A1 Must reject \(N = \frac{1}{2}\)
[7] 
# Question 4:

| Working | Mark | Guidance |
|---------|------|----------|
| $\sum_{r=1}^{N} r^2 = (3N-2)^2$ | | |
| $\frac{1}{6}N(N+1)(2N+1) = 9N^2-12N+4$ | M1 | |
| $2N^3+3N^2+N = 54N^2-72N+24$ | A1 | |
| $2N^3-51N^2+73N-24 = 0$ | A1 | cao |
| Finding one factor, e.g. $(N-1)$: $(N-1)(2N^2-49N+24) = 0$ | B1, m1 | $(N-k)$ form; Linear $\times$ Quadratic (2 terms correct) |
| $\therefore (N-1)(2N-1)(N-24) = 0$ | | |
| $\therefore N=1$ or $N=\frac{1}{2}$ or $N=24$ | A1 | |
| Therefore $N = 1, 24$ | A1 | Must reject $N = \frac{1}{2}$ |
| | [7] | |
4. The positive integer $N$ is such that $1 ^ { 2 } + 2 ^ { 2 } + 3 ^ { 2 } + \ldots + N ^ { 2 } = ( 3 N - 2 ) ^ { 2 }$.

Write down and simplify an equation satisfied by $N$. Hence find the possible values of $N$.\\

\hfill \mbox{\textit{WJEC Further Unit 1 2022 Q4 [7]}}
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