# The result will be an integer

Algebra Level 3

$\large x=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\cdots +\sqrt{1+\frac{1}{2017^2}+\frac{1}{2018^2}}$

Find $$\left \lfloor \sqrt x \right \rfloor$$.


Notation: $$\lfloor \cdot \rfloor$$ denotes the floor function.

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