The sequence a₁, a₂, a₃, ... is such that, for all positive integers n, a_n = (n+5)/√(n²-n+1) - (n+6)/√(n²+n+1). The sum ∑ᴺ_n=1 a_n is denoted by S_N. Find

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2015). The mark or mark type is shown in brackets.

1. [M1] (i) Identify a_n as a difference or recognise telescoping sum nature
2. [A1] (i) Correctly identify the first and last terms of the telescoping sum (e.g., S_N = (6/√1 - (N+6)/√(N²+N+1)))
3. [A1] (i) Calculate S₃₀ = 6 - 36/√931 = 4.820
4. [M1] (ii) Set up inequality 6 - (N+6)/√(N²+N+1) > 4.9
5. [A1] (ii) Rearrange and manipulate to a quadratic inequality (e.g., (N+6)/√(N²+N+1) < 1.1)
6. [dM1] (ii) Solve the resulting quadratic or equivalent inequality for N (dependent on previous M1)
7. [A1] (ii) State the least integer value of N: 55