Edexcel F1 2021 January — Question 5 7 marks

Exam BoardEdexcel
ModuleF1 (Further Pure Mathematics 1)
Year2021
SessionJanuary
Marks7
PaperDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeSum from n+1 to 2n or similar range
DifficultyStandard +0.3 This is a straightforward Further Maths question requiring algebraic manipulation of standard summation formulae. Part (a) involves expanding brackets and applying given formulae (routine technique), while part (b) uses the 'sum from n+1 to 2n' trick of subtracting two sums. Though it requires careful algebra across multiple steps, the techniques are standard for F1 and no novel insight is needed, making it slightly easier than average for Further Maths.
Spec4.06a Summation formulae: sum of r, r^2, r^3

5. (a) Using the formulae for \(\sum _ { r = 1 } ^ { n } r\) and \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\), show that $$\sum _ { r = 1 } ^ { n } ( r + 1 ) ( r + 5 ) = \frac { n } { 6 } ( n + 7 ) ( 2 n + 7 )$$ for all positive integers \(n\).
(b) Hence show that $$\sum _ { r = n + 1 } ^ { 2 n } ( r + 1 ) ( r + 5 ) = \frac { 7 n } { 6 } ( n + 1 ) ( a n + b )$$ where \(a\) and \(b\) are integers to be determined.
VI4V SIHI NI JIIIM ION OCVIAN SIHI NI IHMM I ON OOVAYV SIHI NI JIIIM ION OO

5. (a) Using the formulae for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 2 }$, show that

$$\sum _ { r = 1 } ^ { n } ( r + 1 ) ( r + 5 ) = \frac { n } { 6 } ( n + 7 ) ( 2 n + 7 )$$

for all positive integers $n$.\\
(b) Hence show that

$$\sum _ { r = n + 1 } ^ { 2 n } ( r + 1 ) ( r + 5 ) = \frac { 7 n } { 6 } ( n + 1 ) ( a n + b )$$

where $a$ and $b$ are integers to be determined.\\

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VI4V SIHI NI JIIIM ION OC & VIAN SIHI NI IHMM I ON OO & VAYV SIHI NI JIIIM ION OO \\
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\hfill \mbox{\textit{Edexcel F1 2021 Q5 [7]}}