The sequence a1, a2, a3, ... is such that a₁ > 5 and an+1 = 4an/5 + 5/an for every positive integer n. Prove by mathematical induction that a„ > 5 for every positive integer n. Prove also that an > an+1 for every positive integer n.

Mark scheme

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

1. [B1] a₁ > 5 (given) ⇒ H₁ is true.
2. [B1] Assume Hk is true for some positive integer k, i.e. ak = 5 + δ, where δ > 0.
3. [M1A1] ak+1-5 = 4ak²/5 + 5/ak - 5 = (4ak²+25-25ak)/5ak = (4ak-5)(ak-5)/5ak > 0, ⇒ ak+1 > 5
4. [M1] Or ak+1 = 4/5(5+δ) + 5/(5+δ) = 4 + 4/5δ + 1/(1+δ/5) = 4 + 4/5δ + (1 - δ/5 + δ²/25 - ...) for 0 < δ < 5
5. [A1] = 5 + 3/5δ + O(δ²) ≥ 5, (δ ≥ 5 is trivial).
6. [A1] Hk ⇒ Hk+1 and H₁ is true, hence by mathematical induction, the result is true for all n∈ Z⁺ (N.B. The minimum requirement is 'true for all positive integers'.)
7. [M1] Ak+1-ak = 4ak/5 + 5/ak - ak = 5/ak - ak/5
8. [A1] 5/ak < 1 and ak/5 > 1 ⇒ ak+1 − ak < 0 ⇒ ak+1 < Ak