Edexcel F1 2024 June — Question 7 8 marks

Exam BoardEdexcel
ModuleF1 (Further Pure Mathematics 1)
Year2024
SessionJune
Marks8
PaperDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeFinding n for given sum value
DifficultyStandard +0.3 This is a straightforward Further Maths summation question requiring standard formulas (∑r², ∑r, ∑r³) and algebraic manipulation. Part (a) is routine application of formulas, part (b) requires substitution and solving a cubic equation that likely factors nicely. While it's Further Maths content, it's a standard textbook exercise with no novel insight required, making it slightly easier than average overall.
Spec4.06a Summation formulae: sum of r, r^2, r^3

  1. In this question use the standard results for summations.
    1. Show that for all positive integers \(n\)
    $$\sum _ { r = 1 } ^ { n } \left( 12 r ^ { 2 } + 2 r - 3 \right) = A n ^ { 3 } + B n ^ { 2 }$$ where \(A\) and \(B\) are integers to be determined.
  2. Hence determine the value of \(n\) for which $$\sum _ { r = 1 } ^ { 2 n } r ^ { 3 } - \sum _ { r = 1 } ^ { n } \left( 12 r ^ { 2 } + 2 r - 3 \right) = 270$$

\begin{enumerate}
  \item In this question use the standard results for summations.\\
(a) Show that for all positive integers $n$
\end{enumerate}

$$\sum _ { r = 1 } ^ { n } \left( 12 r ^ { 2 } + 2 r - 3 \right) = A n ^ { 3 } + B n ^ { 2 }$$

where $A$ and $B$ are integers to be determined.\\
(b) Hence determine the value of $n$ for which

$$\sum _ { r = 1 } ^ { 2 n } r ^ { 3 } - \sum _ { r = 1 } ^ { n } \left( 12 r ^ { 2 } + 2 r - 3 \right) = 270$$

\hfill \mbox{\textit{Edexcel F1 2024 Q7 [8]}}