The result will be an integer

Algebra Level 3

x = 1 + 1 1 2 + 1 2 2 + 1 + 1 2 2 + 1 3 2 + ⋯ + 1 + 1 201 7 2 + 1 201 8 2 \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 ⌊ x ⌋ \left \lfloor \sqrt x \right \rfloor .


Notation: ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .


The answer is 44.

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Skye Rzym by SKYE RZYM , 20, Indonesia

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1 solution

x = 1 + 1 1 2 + 1 2 2 + 1 + 1 2 2 + 1 3 2 + 1 + 1 3 2 + 1 4 2 + . . . + 1 + 1 201 7 2 + 1 201 8 2 = ∑ n = 1 2017 1 + 1 n 2 + 1 ( n + 1 ) 2 = ∑ n = 1 2017 n 2 ( n + 1 ) 2 + ( n + 1 ) 2 + n 2 n 2 ( n + 1 ) 2 = ∑ n = 1 2017 n 4 + 2 n 3 + 3 n 2 + 2 n + 1 n 2 ( n + 1 ) 2 = ∑ n = 1 2017 ( n 2 + n + 1 ) 2 n 2 ( n + 1 ) 2 = ∑ n = 1 2017 n 2 + n + 1 n ( n + 1 ) = ∑ n = 1 2017 ( 1 + 1 n ( n + 1 ) ) = ∑ n = 1 2017 ( 1 + 1 n − 1 n + 1 ) = 2017 + 1 1 − 1 2018 ≈ 2018 \begin{aligned} x & = \sqrt{1+\frac 1{1^2} + \frac 1{2^2}} + \sqrt{1+\frac 1{2^2} + \frac 1{3^2}} + \sqrt{1+\frac 1{3^2} + \frac 1{4^2}} + ... + \sqrt{1+\frac 1{2017^2} + \frac 1{2018^2}} \\ & = \sum_{n=1}^{2017} \sqrt{1+\frac 1{n^2} + \frac 1{(n+1)^2}} \\ & = \sum_{n=1}^{2017} \sqrt{\frac {n^2(n+1)^2 + (n+1)^2 +n^2}{n^2(n+1)^2}} \\ & = \sum_{n=1}^{2017} \sqrt{\frac {n^4+2n^3+3n^2+2n+1}{n^2(n+1)^2}} \\ & = \sum_{n=1}^{2017} \sqrt{\frac {(n^2+n+1)^2}{n^2(n+1)^2}} \\ & = \sum_{n=1}^{2017} \frac {n^2+n+1}{n(n+1)} \\ & = \sum_{n=1}^{2017} \left(1 + \frac 1{n(n+1)}\right) \\ & = \sum_{n=1}^{2017} \left(1 + \frac 1n - \frac 1{n+1} \right) \\ & = 2017 + \frac 11 - \frac 1{2018} \\ & \approx 2018 \end{aligned}

⟹ ⌊ x ⌋ = ⌊ 2018 ⌋ = 44 \implies \left \lfloor \sqrt x \right \rfloor = \left \lfloor 2018 \right \rfloor = \boxed{44}

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did the same except factrosation i did putting of values although question if from last year NDA paper

Nivedit Jain - 4 years, 3 months ago

Did the same way

I Gede Arya Raditya Parameswara - 4 years, 3 months ago

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