Olympiad Algebra Problem:
If $a,\,b,\,x,\,y\in R$ such that
$ax +by =7$
$ax^2+by^2=49$
$ax^3+by^3=133$
$ax^4+by^4=406$
Evaluate $ax^5+by^5$.
The solution from the U.K. mathematician:
Showing posts with label thinking outside the box. Show all posts
Showing posts with label thinking outside the box. Show all posts
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