Use the method of differences to show that Σ_{r=1}^N 1/((2r + 1)(2r + 3)) = 1/6 - 1/(2(2N + 3)). Deduce that Σ_{r=N+1}^2N 1/((2r + 1)(2r + 3)) < 1/(8N).
- (a)Use the method of differences to show that Σ_{r=1}^N 1/((2r + 1)(2r + 3)) = 1/6 - 1/(2(2N + 3)).[5]
- (b)Deduce that Σ_{r=N+1}^2N 1/((2r + 1)(2r + 3)) < 1/(8N).[4]
Mark scheme
Marking points from the official mark scheme (May/June 2013). The mark or mark type is shown in brackets.
- M1A1 Finds partial fractions. 1/((2r+1)(2r+3)) = 1/2 * (1/(2r+1) - 1/(2r+3))
- M1A1 Expresses terms as differences. = 1/2 [(1/3 - 1/5) + ... + (1/(2N+1) - 1/(2N+3))]
- A1 Shows cancellation. = 1/2 [1/3 - 1/(2N+3)] (AG)
- M1 Uses Σ_{r=N+1}^2N = Σ_{r=1}^2N - Σ_{r=1}^N
- A1 Applies result = 1/2 [1/(2N+3) - 1/(4N+3)]
- M1 and simplifies. N/((2N+3)(4N+3))
- A1 Deduces inequality. < 1/(2N.4N) = 1/(8N) (AG)
Write your answer and get it marked
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.
Sign up free →