The result will be an integer

Algebra Level pending

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

Find \(\left \lfloor \sqrt x \right \rfloor\).

\(\)
Notation: \( \lfloor \cdot \rfloor \) denotes the floor function.

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