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Algebra Level 5

Given that f ( x ) = 2 5 x 2 5 x + 5 f(x) = \dfrac{25^x}{25^x + 5} . If the value of the sum, ∑ a = 1 2015 f ( a 2016 ) \displaystyle \sum_{a=1}^{2015} f \left(\dfrac a{2016} \right) is equal to A B \dfrac AB , where A A and B B are coprime positive integers, find the value of A + B A+B .

Hint : Generalize this.


The answer is 2017.

Aditya Kumar by Aditya Kumar , 22, India

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2 solutions

Tanishq Varshney
Dec 24, 2015

Clearly

f ( x ) + f ( 1 − x ) = 1 f(x)+f(1-x)=1

Let S = ∑ a = 1 2015 f ( a / 2016 ) S=\sum_{a=1}^{2015}f(a/2016)

Using ∑ r = a b f ( r ) = ∑ r = a b f ( a + b − r ) \large{\sum^{b}_{r=a}f(r)=\sum^{b}_{r=a}f(a+b-r)}

2 S = ∑ a = 1 2015 1 \large{2S=\sum^{2015}_{a=1}1} ∵ f ( a / 2016 ) + f ( 1 − a / 2016 ) = 1 \quad \quad \because f(a/2016)+f(1-a/2016)=1

S = 2015 2 \large{S=\frac{2015}{2}}

19 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice observation

Mardokay Mosazghi - 5 years, 5 months ago

Did by this way but @Tanishq Varshney , shall we have to show that since the no. Of terms are odd ,therefore there must be a middle term that is required to be explicitly written( that is f(1008/2016) )and we also have to show that it's value equal to 1/2 ...am I right?

Righved K - 5 years, 5 months ago

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Yes you are right

Department 8 - 5 years, 5 months ago

I also did by same method. One can only apply this method when he had done this before

Priyanshu Mishra - 5 years, 5 months ago

Nice change of the sum of f ( r ) f(r) to f ( a + b − r ) f(a+b-r) . You did not answered my question Are you going to WWE LiveIndia? Please see this

Department 8 - 5 years, 5 months ago
William Isoroku
Dec 26, 2015

Hint: rewrite the equation as

f ( x ) = 1 1 + 2 5 1 2 − x f(x)=\frac{1}{1+25^{\frac{1}{2}-x}}

Then the summation will be much easier because all of the terms except for the middle term, which is 1 2 \frac{1}{2} , will add to 1 1 . There will be 1007 terms, excluding the middle term, so the sum will be 1007 ( 1 ) + 1 2 = 2015 2 1007(1)+\frac{1}{2}=\frac{2015}{2}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

I liked your method :)

Aditya Kumar - 5 years, 5 months ago

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