Find the value

Algebra Level 3

A = f ( 1 100 ) + f ( 2 100 ) + f ( 3 100 ) + . . . . . . + f ( 99 100 ) + f ( 100 100 ) + f ( 100 99 ) + . . . . . . + f ( 100 3 ) + f ( 100 2 ) + f ( 100 1 ) \begin{aligned} A & = f \left(\frac{1}{100}\right) + f\left(\frac{2}{100}\right) + f\left(\frac{3}{100}\right) + ... \\ & \quad \quad \quad ... + f\left(\frac{99}{100}\right) + f\left(\frac{100}{100}\right) + f\left(\frac{100}{99}\right) + ... \\ & \quad \quad \quad \quad \quad \quad ... + f\left(\frac {100}3 \right) + f\left(\frac{100}{2}\right) + f\left(\frac{100}{1}\right) \end{aligned}

Given that f ( x ) = x 2 1 + x 2 f(x) = \dfrac {x^2}{1+x^2} , find the value of A A above.


The answer is 99.5.

Manos Garefalakis by Manos Garefalakis , 33, Greece

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1 solution

Consider the following, where k k is a positive integer.

{ f ( k 100 ) = k 2 10 0 2 1 + k 2 10 0 2 = k 2 10 0 2 + k 2 f ( 100 k ) = 10 0 2 k 2 1 + 10 0 2 k 2 = 10 0 2 k 2 + 10 0 2 ⟹ f ( k 100 ) + f ( 100 k ) = k 2 10 0 2 + k 2 + 10 0 2 k 2 + 10 0 2 = 1 \begin{cases} f \left(\dfrac k{100}\right) = \dfrac {\frac {k^2}{100^2}}{1+\frac {k^2}{100^2}} = \dfrac {k^2}{100^2+k^2} \\ f \left(\dfrac {100}k\right) = \dfrac {\frac {100^2}{k^2}}{1+\frac {100^2}{k^2}} = \dfrac {100^2}{k^2+100^2} \end{cases} \implies f \left(\dfrac k{100}\right) + f \left(\dfrac {100}k\right) = \dfrac {k^2}{100^2+k^2} + \dfrac {100^2}{k^2+100^2} =1

⟹ A = ∑ k = 1 99 f ( k 100 ) + f ( 100 100 ) + ∑ k = 1 99 f ( 100 k ) = ∑ k = 1 99 ( f ( k 100 ) + f ( 100 k ) ) + f ( 1 ) = ∑ k = 1 99 1 + 1 2 1 + 1 2 = 99 + 1 2 = 99.5 \begin{aligned} \implies A & = \sum_{k=1}^{99} f \left(\frac k{100}\right) + f \left(\frac {100}{100}\right) + \sum_{k=1}^{99} f \left(\frac {100}k\right) \\ & = \sum_{k=1}^{99} \left(f \left(\frac k{100}\right) + f \left(\frac {100}k \right)\right) + f(1) \\ & = \sum_{k=1}^{99} 1 + \frac {1^2}{1+1^2} \\ & = 99 + \frac 12 = \boxed{99.5} \end{aligned}

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