ik snap een bepaalde stap niet die ze in het boek doen:
Let
\(x_{1}=1\)
and \(x_{2}\)
, for \(n\geq 2\)
, let \(x_{n+1} = x_{n} + 2x_{n-1}\)
. Prove that \(x_{n}\)
is divisible by 3 if and only if n is divisible by 3.This is one of the examples we promised in Section 3.1, where the key to the proof is a clever choice of S. Before proceeding, you might try some logical choices of S and see if proofs can be constructed using these choices.
Let
\(S = \{ n \in \mathbb{N} : x_{n+6} - x_{n} \mbox{is divisible by } 3 \}\)
Since \(x_{7} - x_{1} = 43 - 1\)
and \(x_{8} - x_{2} = 85 - 1 \)
, then 1,2 \(\in\)
S. Let \(n \geq 2\)
and assume that \({1,2,...,n}\subseteq S\)
. Then\(x_{n+1+6} - x_{n+1} = x_{n+7} - x_{n+1}\)
\(x_{n+1+6} - x_{n+1} = x_{n+6} + 2x_{n+5} - x_{n} -2x_{n-1}\)
\(x_{n+1+6} - x_{n+1} = x_{n+6} - x_{n} + 2(x_{k+6} - x_{k})\)
where \(k=n=-1\)
. Since both k and n belong to S, both \(2(x_{k+6} - x_{k})\)
and \(x_{n+6}-x_{n}\)
are divisible by 3, and so \(n+1 \in S\)
. By the Second Principle Mathematical, \(S=\mathbb{N}\)
, and it follows from this equality that \(x_{n}\)
is divisible by 3 if and only if n is divisible by 3.(Mogelijk zijn er tikfouten, mijn excuses daarvoor alvast)
Mijn vraag is, hoe komen ze van
\(x_{n+1} = x_{n} + 2x_{n-1}\)
naar \( x_{n+1+6}-x_{n+1}\)
ik zie deze "overgang" namelijk niet echt. (het komt voor mij zomaar uit de lucht vallen).Bedankt voor uw uitleg alvast.
