1976 IMO Problems/Problem 6
Problem
A sequence is defined by
Prove that for any positive integer we have
(where denotes the smallest integer )
Solution
Let the sequence be defined as \[ x_{0}=0,x_{1}=1, x_{n}=x_{n-1}+2x_{n-2} \] We notice Because the roots of the characteristic polynomial are and . \\newline We also see , We want to prove This is done by induction \subsection*{Base case} For ses det
\subsection*{Induction step} Assume We notice \begin{align*} 2^{x_{n}-2x_{n-1}}+2^{-x_{n}+2x_{n-1}} &=2^{x_{n-1}+2x_{n-2}-2x_{n-1}}+2^{-(x_{n-1}+2x_{n-2})+2x_{n-1}}\\ &=2^{-x_{n-1}+2x_{n-2}}+2^{x_{n-1}-2x_{n-2}}\\ &=2^{x_1}+2^{-x_1} \end{align*}\newline We then want to show This can be done using induction \subsection*{Base case} For , it is clear that and Therefore, the base case is proved \subsection*{Induktionsskridt} Assume for all natural at \newline Then we have that: \begin{align*} a_n &= a_{n}(a_{n-1}^{2}-2)-a_{1} \\
&= (2^{x_{n-1}}+2^{-x_{n-1}})((2^{x_{n-2}}+2^{-x_{n-2}})^2-2)-(2^{x_{1}}+2^{-x_{0}}) \\ &= (2^{x_{n-1}}+2^{-x_{n-1}})((2^{x_{n-2}})^2+(2^{-x_{n-2}})^2+2*2^{x_{n-2}}*2^{-x_{n-2}}-2)-(2^{x_{1}}+2^{-x_{0}}) \\ &=(2^{x_{n-1}}+2^{-x_{n-1}})((2^{2x_{n-2}})+(2^{-2x_{n-2}})+2*2^{x_{n-2}-x_{n-2}}-2)-(2^{x_{1}}+2^{-x_{0}}) \\ &=(2^{x_{n-1}}+2^{-x_{n-1}})((2^{2x_{n-2}})+(2^{-2x_{n-2}})+2-2)-(2^{x_{1}}+2^{-x_{0}}) \\
&=2^{x_{n-1}}*2^{2x_{n-2}}+2^{-x_{n-1}}2^{-2x_{n-2}}+2^{x_{n-1}}*2^{-2x_{n-2}}+2^{-x_{n-1}}*2^{2x_{n-2}}-2^{x_{1}}-2^{-x_{0}}\\ &=2^{x_{n-1}+2x_{n-2}}+2^{-(x_{n-1}+2x_{n-2}})+2^{x_{n-1}-2x_{n-2}}+2^{-x_{n-1}+2x_{n-2}}-2^{x_{1}}-2^{-x_{0}}\\ &=2^{x_{n}}+2^{-x_{n}}+2^{x_{n-1}-2x_{n-2}}+2^{-x_{n-1}+2x_{n-2}}-2^{x_{1}}-2^{-x_{0}} \end{align*} From our first induction proof we have that: Then: We notice , Because and , for all Finally we conclude
See also
1976 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Final Question |
All IMO Problems and Solutions |