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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, 3 months ago

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Hint :- \( \phi^2 = \phi + 1 \) Siddhartha Srivastava · 2 years, 3 months ago

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@Siddhartha Srivastava

Samuraiwarm Tsunayoshi · 2 years, 3 months ago

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Is the answer n*(n+1)/2 ??? Abhineet Nayyar · 2 years, 3 months ago

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@Abhineet Nayyar Yup Samuraiwarm Tsunayoshi · 2 years, 3 months ago

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