A-LevelFurther MathematicsSummation of series9231/12 · Oct/Nov 2023 · Q19 Marks

Use standard results from the list of formulae (MF19) to find \sum_{r=1}^{n} (3r^2 +3r+1) in terms of n, simplifying your answer.

(a)Use standard results from the list of formulae (MF19) to find \sum_{r=1}^{n} (3r^2 +3r+1) in terms of n, simplifying your answer.[3]
(b)Show that 1/(r³) - 1/((r+1)³) = (3r² +3r+1)/(r³ (r+1)³) and hence use the method of differences to find Σᵣ₌₁ⁿ (3r² +3r+1)/(r³ (r+1)³)[5]
(c)Deduce the value of Σᵣ₌₁^(∞) (3r² +3r+1)/(r³ (r+1)³)[1]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

[M1 A1] Substitutes correct formulae from MF19. ½n(n+1)(2n+1)+n(n + 1) + n
[A1] Simplifies n³ +3n² + 3n
[M1 A1] Puts over a common denominator and expands, AG. ((r+1)³ - r³)/(r³(r+1)³) = (3r²+3r+1)/(r³(r+1)³)
[M1 A1] Shows three complete terms, including last. Σᵣ₌₁ⁿ (1/(r³) - 1/((r+1)³)) = (1/(1³) - 1/(2³)) + (1/(2³) - 1/(3³)) + … + (1/(n³) - 1/((n+1)³)) = 1 - 1/((n+1)³)
[A1] 1 - 1/((n+1)³)
[B1FT] FT from their answer to part (b). 1

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.

About this question

Topic

Summation of series, A-Level Further Mathematics (syllabus 9231). Structured question, 9 marks.

Source

A-Level Further Mathematics, Oct/Nov 2023 Paper 1 Variant 2, question 1.

Questions and mark schemes are from past examination papers and remain the copyright of the exam board. How Oracle Prep builds its content