Difference between revisions of "1976 IMO Problems/Problem 6"

(Problem)
Line 6: Line 6:
 
Prove that for any positive integer <math>n</math> we have
 
Prove that for any positive integer <math>n</math> we have
  
<cmath>[u_{n}] = 2^{\frac {(2^{n} - ( - 1)^{n})}{3}}</cmath>
+
<cmath>\lfloor u_{n} \rfloor = 2^{\frac {(2^{n} - ( - 1)^{n})}{3}}</cmath>
  
(where [x] denotes the smallest integer <math>\leq</math> x)<math>.</math>
+
(where <math>\lfloor x\rfloor</math> denotes the smallest integer <math>\leq</math> x)<math>.</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 09:47, 26 February 2008

Problem

A sequence $(u_{n})$ is defined by

\[u_{0} = 2 \quad u_{1} = \frac {5}{2}, u_{n + 1} = u_{n}(u_{n - 1}^{2} - 2) - u_{1} \quad \textnormal{for} n = 1,\ldots\]

Prove that for any positive integer $n$ we have

\[\lfloor u_{n} \rfloor = 2^{\frac {(2^{n} - ( - 1)^{n})}{3}}\]

(where $\lfloor x\rfloor$ denotes the smallest integer $\leq$ x)$.$

Solution

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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