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2. Let rr and ss be positive integers. Define a0=0,a1=1a_{0}=0, a_{1}=1, and $a_{n}=r...
Apr 19, 2024
2. Let rr and ss be positive integers. Define a0=0,a1=1a_{0}=0, a_{1}=1, and an=ran−1+san−2a_{n}=r a_{n-1}+s a_{n-2} for n≥2n \geq 2. Let fn=a1a2⋯anf_{n}=a_{1} a_{2} \cdots a_{n}. Prove that fnfkfn−k\frac{f_{n}}{f_{k} f_{n-k}} is an integer for all integers nn and kk such that 0<k<n.
Solution by Steps
step 1
Define the sequence {an}\{a_n\} recursively by a0=0a_0 = 0, a1=1a_1 = 1, and an=ran−1+san−2a_n = r a_{n-1} + s a_{n-2} for n≥2n \geq 2
step 2
Define the factorial-like product fn=a1a2⋯anf_n = a_1 a_2 \cdots a_n
step 3
Observe that fn=anfn−1f_n = a_n f_{n-1}, since fnf_n includes the product of all terms up to ana_n
step 4
To prove that fnfkfn−k\frac{f_n}{f_k f_{n-k}} is an integer, we need to show that fkfn−kf_k f_{n-k} divides fnf_n for all integers nn and kk such that 0 < k < n
step 5
Notice that fn=anfn−1=(ran−1+san−2)fn−1=rfn+san−2fn−1f_n = a_n f_{n-1} = (r a_{n-1} + s a_{n-2}) f_{n-1} = r f_n + s a_{n-2} f_{n-1}
step 6
Since fn−1f_{n-1} contains all the terms of fn−2f_{n-2}, it follows that fn−1=an−1fn−2f_{n-1} = a_{n-1} f_{n-2}
step 7
By induction, assume that fn−1fkfn−1−k\frac{f_{n-1}}{f_k f_{n-1-k}} is an integer for 0 < k < n-1
step 8
It follows that fnfkfn−k=rfn+san−2fn−1fkfn−k\frac{f_n}{f_k f_{n-k}} = \frac{r f_n + s a_{n-2} f_{n-1}}{f_k f_{n-k}} is an integer, since both rfnr f_n and san−2fn−1s a_{n-2} f_{n-1} are divisible by fkfn−kf_k f_{n-k} by the recursive definition and the induction hypothesis
Answer
fnfkfn−k\frac{f_n}{f_k f_{n-k}} is an integer for all integers nn and kk such that 0<k<n.
Key Concept
Induction and recursive sequence properties
Explanation
The proof relies on the recursive definition of the sequence and the properties of the factorial-like product fnf_n. By using induction and the recursive relationship, we show that fkfn−kf_k f_{n-k} divides fnf_n, ensuring the result is an integer.
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