WJEC Further Unit 1 2019 June — Question 7 8 marks

Exam BoardWJEC
ModuleFurther Unit 1 (Further Unit 1)
Year2019
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeStandard summation formulae application
DifficultyModerate -0.3 This is a straightforward application of standard summation formulae requiring algebraic manipulation. Part (a) involves expanding (r+2)² and applying Σr² and Σr formulae, then factorizing into the given form. Part (b) is direct substitution. While it requires careful algebra, it's a routine Further Maths exercise with no novel problem-solving or insight needed.
Spec4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series
7. (a) Find an expression for \(\sum _ { r = 1 } ^ { 2 m } ( r + 2 ) ^ { 2 }\) in the form \(\frac { 1 } { 3 } m \left( a m ^ { 2 } + b m + c \right)\), where \(a , b , c\) are integers whose values are to be determined.
(b) Hence, calculate \(\sum _ { r = 1 } ^ { 20 } ( r + 2 ) ^ { 2 }\).
Part (a):
AnswerMarks Guidance
\(\sum_{r=1}^{2m} (r^2 + 4r + 4)\)M1
Use of formulae for \(\sum r^2\) and \(\sum r\) and \(\sum 4\): \(= \frac{1}{6}(2m)(2m+1)(4m+1) + 4 \times \frac{1}{2}(2m)(2m+1) + 8m\)M1, A1
\(= \frac{1}{6}m(16m^2 + 12m + 2 + 48m + 24 + 48) = \frac{1}{6}m(8m^2 + 30m + 37)\)A1 cao
Part (b):
AnswerMarks Guidance
Substituting appropriate values into their expressions in (a) i.e. \(m = 5\) and \(m = 10\) for \(2m\)M1 FT (a)
Subtracting values for \(r = 10\) from \(r = 20\): \(\sum_{r=1}^{20}(r+2)^2 - \sum_{r=1}^{10}(r+2)^2 = \frac{1}{3} \times 10 \times (8 \times 100 + 30 \times 10 + 37) - \frac{1}{3} \times 5 \times (8 \times 25 + 30 \times 5 + 37) = 3145\)m1, A1, A1 cao
If candidate has used \(\sum_{r=1}^{m}(r+2)^2\) expression obtained is \(= \frac{1}{6}m(2m^2 + 15m + 37)\). Then calculations are: Subtracting values for \(r = 10\) from \(r = 20\): \(\sum_{r=1}^{20}(r+2)^2 - \sum_{r=1}^{10}(r+2)^2 = \frac{1}{6} \times 20 \times (2 \times 400 + 15 \times 20 + 37) - \frac{1}{6} \times 10 \times (2 \times 100 + 15 \times 10 + 37) = 3145\)(M1), (m1), (A1), (A1) FT (a); cao 
**Part (a):**
$\sum_{r=1}^{2m} (r^2 + 4r + 4)$ | M1 |

Use of formulae for $\sum r^2$ and $\sum r$ and $\sum 4$: $= \frac{1}{6}(2m)(2m+1)(4m+1) + 4 \times \frac{1}{2}(2m)(2m+1) + 8m$ | M1, A1 |

$= \frac{1}{6}m(16m^2 + 12m + 2 + 48m + 24 + 48) = \frac{1}{6}m(8m^2 + 30m + 37)$ | A1 | cao

**Part (b):**
Substituting appropriate values into their expressions in (a) i.e. $m = 5$ and $m = 10$ for $2m$ | M1 | FT (a)

Subtracting values for $r = 10$ from $r = 20$: $\sum_{r=1}^{20}(r+2)^2 - \sum_{r=1}^{10}(r+2)^2 = \frac{1}{3} \times 10 \times (8 \times 100 + 30 \times 10 + 37) - \frac{1}{3} \times 5 \times (8 \times 25 + 30 \times 5 + 37) = 3145$ | m1, A1, A1 | cao

If candidate has used $\sum_{r=1}^{m}(r+2)^2$ expression obtained is $= \frac{1}{6}m(2m^2 + 15m + 37)$. Then calculations are: Subtracting values for $r = 10$ from $r = 20$: $\sum_{r=1}^{20}(r+2)^2 - \sum_{r=1}^{10}(r+2)^2 = \frac{1}{6} \times 20 \times (2 \times 400 + 15 \times 20 + 37) - \frac{1}{6} \times 10 \times (2 \times 100 + 15 \times 10 + 37) = 3145$ | (M1), (m1), (A1), (A1) | FT (a); cao

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7. (a) Find an expression for $\sum _ { r = 1 } ^ { 2 m } ( r + 2 ) ^ { 2 }$ in the form $\frac { 1 } { 3 } m \left( a m ^ { 2 } + b m + c \right)$, where $a , b , c$ are integers whose values are to be determined.\\
(b) Hence, calculate $\sum _ { r = 1 } ^ { 20 } ( r + 2 ) ^ { 2 }$.\\

\hfill \mbox{\textit{WJEC Further Unit 1 2019 Q7 [8]}}
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