Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Prove it Ukraine

Sequence (an)(a_n) is defined as follows:
a1=1, a2=2, an+2=(n+1)(an+an+1) a_1 = 1,\ a_2 = 2,\ a_{n+2} = (n+1)(a_n + a_{n+1})
for each natural nn. How many zeros does a2011a_{2011} end with?

Solution

We first prove that an=n!a_n = n! by PMI:
an+2=(n+1)(n!+(n+1)!)=n!(n+1)(n+2)=(n+2)! a_{n+2} = (n+1)(n! + (n+1)!) = n!(n+1)(n+2) = (n+2)!
One can now find the number of zeros using the well-known formula:
[20115]+[201152]+[201153]+⋯=402+80+16+3=501. \left[ \frac{2011}{5} \right] + \left[ \frac{2011}{5^2} \right] + \left[ \frac{2011}{5^3} \right] + \dots = 402 + 80 + 16 + 3 = 501.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.