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!

#Algebra #NumberTheory #Summation #GoldenRatio #FloorFunction

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
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Comments

Hint :- $\phi^2 = \phi + 1$

Siddhartha Srivastava - 6 years, 5 months ago

Samuraiwarm Tsunayoshi - 6 years, 5 months ago

Is the answer n*(n+1)/2 ???

A Former Brilliant Member - 6 years, 5 months ago

Yup

Samuraiwarm Tsunayoshi - 6 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$