Do you know the future

Algebra Level pending

Let f ( x ) = a x a x + a , a > 0 f(x)=\frac{a^x}{a^x+\sqrt{a}}, a>0

Find the value of i = 1 2016 2 f ( i 2017 ) \sum_{i=1}^{2016}2f(\frac{i}{2017})


The answer is 2016.

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1 solution

Gautam Jha
Apr 18, 2015

Here, f ( x ) = a x a x + a f(x)=\frac{a^x}{a^x+\sqrt{a}}

and, f ( 1 x ) = a 1 x a 1 x + a f(1-x)=\frac{a^{1-x}}{a^{1-x}+\sqrt{a}}

Note, that f ( x ) + f ( 1 x ) = 1 f(x)+f(1-x)=1

2[ f ( 1 2017 ) + f ( 2 2017 ) + . . . . . . . . . . . . . + f ( 2016 2017 ) ] f(\frac{1}{2017})+f(\frac{2}{2017})+.............+f(\frac{2016}{2017})]

= 2 [ f ( 1 2017 ) + f ( 2016 2017 ) + f ( 2 2017 ) + f ( 2015 2017 ) + . . . . . . + f ( 1008 2017 ) + f ( 1009 2017 ) ] =2[f(\frac{1}{2017})+f(\frac{2016}{2017})+f(\frac{2}{2017})+f(\frac{2015}{2017})+......+f(\frac{1008}{2017})+f(\frac{1009}{2017})]

= 2 [ f ( 1 2017 ) + f ( 1 1 2017 ) + f ( 2 2017 ) + f ( 1 2 2017 ) + . . . . . . + f ( 1008 2017 ) + f ( 1 1008 2017 ) ] =2[f(\frac{1}{2017})+f(1-\frac{1}{2017})+f(\frac{2}{2017})+f(1-\frac{2}{2017})+......+f(\frac{1008}{2017})+f(1-\frac{1008}{2017})]

= 2 [ 1 + 1 + 1........ 2016 2 t i m e s ] =2[1+1+1........\frac{2016}{2} times]

2016 \boxed{2016}

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