22nd IMO 1981 shortlist

A sequence a_n is defined as follows, a₀ = 1, a_n+1 = (1 + 4a_n + √(1 + 24a_n))/16 for n ≥ 0. Find an explicit formula for a_n.

Answer

(1/3)(1 + 1/2ⁿ)(1 + 1/2ⁿ⁺¹)

Solution

Put b_n = 2²ⁿ⁺¹a_n. Then the relation becomes b_n+1 = 2^2n-1 + b_n + 2ⁿ √(2^2n-2 + 3b_n). Put c_n = √(2^2n-2 + 3b_n) and this becomes c_n+1² = c_n² + 3·2ⁿc_n + 9·2^2n-2 = (c_n + 3·2^n-1)². Hence c_n+1 = c_n + 3·2^n-1. Iterating, c_n+1 = 3·2^n-1 + 3·2^n-2 + ... + 3·2⁰ + c₁ = 3(2ⁿ-1) + c₁ = 3·2ⁿ + 1 (we have a₁ = 5/8, so b₁ = 5, c₁ = 4). Hence b_n = (2²ⁿ⁺¹ + 3·2ⁿ + 1)/3, a_n = 1/3 + 1/2ⁿ⁺¹ + 1/(3·2²ⁿ⁺¹) = (1/3)(1 + 1/2ⁿ)(1 + 1/2ⁿ⁺¹).

Thanks to Suat Namli

24th IMO shortlist 1983

© John Scholes
jscholes@kalva.demon.co.uk
12 February 2004
Last corrected/updated 12 Feb 04