Calculating the Sum of f(x) from 0 to 2016

[lfdahl]

Let

\[f(x) = \frac{a^{2x}}{a^{2x}+a}, \;\;\; a \in \Bbb{N}.\]Find the sum:\[ \sum_{j=0}^{2016}f \left ( \frac{j}{2016} \right )\]

 

[Greg]

Let

\[f(x) = \frac{a^{2x}}{a^{2x}+a}, \;\;\; a \in \Bbb{N}.\]Find the sum:\[ \sum_{j=0}^{2016}f \left ( \frac{j}{2016} \right )\]

Spoiler

You can write it as $1-\dfrac{1}{a^{2x-1}+1}$

First term: $1-\dfrac{a}{1+a}$

Last term: $1-\dfrac{1}{1+a}$

Second term: $1-\dfrac{a^{1007/1008}}{a^{1007/1008}+1}$

Second-to-last term: $1-\dfrac{1}{a^{1007/1008}+1}$

It "telescopes" and the middle term is $\dfrac12$, so the sum is $\dfrac{2017}{2}$.

 

[Albert1]

Spoiler

You can write it as $1-\dfrac{1}{a^{2x-1}+1}$

First term: $1-\dfrac{a}{1+a}$

Last term: $1-\dfrac{1}{1+a}$

Second term: $1-\dfrac{a^{1007/1008}}{a^{1007/1008}+1}$

Second-to-last term: $1-\dfrac{1}{a^{1007/1008}+1}$

It "telescopes" and the middle term is $\dfrac12$, so the sum is $\dfrac{2017}{2}$.

marvelous !

 

[lfdahl]

Spoiler

You can write it as $1-\dfrac{1}{a^{2x-1}+1}$

First term: $1-\dfrac{a}{1+a}$

Last term: $1-\dfrac{1}{1+a}$

Second term: $1-\dfrac{a^{1007/1008}}{a^{1007/1008}+1}$

Second-to-last term: $1-\dfrac{1}{a^{1007/1008}+1}$

It "telescopes" and the middle term is $\dfrac12$, so the sum is $\dfrac{2017}{2}$.

Well done, greg1313, thankyou very much for your solution!