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# Golden ratio sum, and squares?! (help mehhh)

Let $$\phi$$ be the golden ratio constant, find the value in terms of $$n$$ of

$\sum\limits_{k=1}^{n} \left(\lfloor k\phi^{2} \rfloor - \lfloor k\phi \rfloor \right)$

I wonder why is this problem called "easy" for someone, but not me!

Note by Samuraiwarm Tsunayoshi
2 years, 6 months ago

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Hint :- $$\phi^2 = \phi + 1$$ · 2 years, 6 months ago

· 2 years, 6 months ago

Is the answer n*(n+1)/2 ??? · 2 years, 6 months ago