Edexcel F1 2024 January — Question 8 8 marks

Exam BoardEdexcel
ModuleF1 (Further Pure Mathematics 1)
Year2024
SessionJanuary
Marks8
PaperDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeSum from n+1 to 2n or similar range
DifficultyChallenging +1.2 Part (a) is routine application of standard summation formulas (Σr, Σr², Σr³) requiring algebraic manipulation but no insight. Part (b) requires the telescoping technique (sum from n to 2n equals sum to 2n minus sum to n-1) and factorization of a quartic expression, which is more challenging than typical but still a standard Further Maths technique with clear methodology.
Spec1.04g Sigma notation: for sums of series4.06a Summation formulae: sum of r, r^2, r^3

  1. (a) Use the standard results for summations to show that, for all positive integers \(n\),
$$\sum _ { r = 1 } ^ { n } r \left( 2 r ^ { 2 } - 3 r - 1 \right) = \frac { 1 } { 2 } n ( n + 1 ) ^ { 2 } ( n - 2 )$$ (b) Hence show that, for all positive integers \(n\), $$\sum _ { r = n } ^ { 2 n } r \left( 2 r ^ { 2 } - 3 r - 1 \right) = \frac { 1 } { 2 } n ( n - 1 ) ( a n + b ) ( c n + d )$$ where \(a\), \(b\), \(c\) and \(d\) are integers to be determined.

\begin{enumerate}
  \item (a) Use the standard results for summations to show that, for all positive integers $n$,
\end{enumerate}

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

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

$$\sum _ { r = n } ^ { 2 n } r \left( 2 r ^ { 2 } - 3 r - 1 \right) = \frac { 1 } { 2 } n ( n - 1 ) ( a n + b ) ( c n + d )$$

where $a$, $b$, $c$ and $d$ are integers to be determined.

\hfill \mbox{\textit{Edexcel F1 2024 Q8 [8]}}