A-LevelFurther MathematicsSummation of series9231/12 · May/June 2018 · Q58 Marks
Let Sₙ = ∑ⁿᵣ₌₁(−1)ʳ⁻¹r².
- (i)Use the standard result for ∑ⁿᵣ₌₁r² given in the List of Formulae (MF10) to show that S₂ⁿ = −n(2n + 1).[4]
- (ii)State the value of limₙ→∞ S₂ⁿ/n² and find limₙ→∞ S₂ⁿ⁺¹/n².[4]
Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- M1 (i) Uses correct difference. S₂ⁿ = 1²-2² +3² -4² + ... + (2n-1)² - (2n)²
- A1 (i) S₂ⁿ = Σ^n_r=1 (2r-1)² - Σ^n_r=1 (2r)²
- M1 (i) S₂ⁿ = Σ^n_r=1 ((2r-1)² - (2r)²) = Σ^n_r=1 (-4r + 1)
- A1 (i) = -n(2n+1) (AG, requires correct working)
- B1 (ii) State the value of limₙ→∞ S₂ⁿ/n² = -2
- M1 (ii) S₂ⁿ⁺¹ = S₂ⁿ + (-1)^(2n) (2n+1)²
- M1 (ii) S₂ⁿ⁺¹ = -n(2n+1) + (2n+1)² = (2n+1)(n+1)
- A1 (ii) Thus limₙ→∞ S₂ⁿ⁺¹/n² = 2
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