Difference between revisions of "1976 IMO Problems/Problem 6"
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A sequence <math>(u_{n})</math> is defined by | A sequence <math>(u_{n})</math> is defined by | ||
− | <cmath>u_{0} = 2 \quad u_{1} = \frac {5}{2}, u_{n + 1} = u_{n}(u_{n - 1}^{2} - 2) - u_{1} \quad \ | + | <cmath>u_{0} = 2 \quad u_{1} = \frac {5}{2}, u_{n + 1} = u_{n}(u_{n - 1}^{2} - 2) - u_{1} \quad \text{ for } n = 1,\cdots</cmath> |
Prove that for any positive integer <math>n</math> we have | Prove that for any positive integer <math>n</math> we have |
Revision as of 16:29, 29 January 2021
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
Base Case:
For ses det
Inductive step:
Assume
We notice
We then want to show
This can be done using induction
Base Case
For , it is clear that
and
Therefore, the base case is proved.
Inductive Step
Assume for all natural at
\newline
Then we have that:
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 |