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

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The discussion focuses on calculating the sum of the function f(x) defined as f(x) = a^(2x) / (a^(2x) + a) for a natural number a, evaluated from j=0 to 2016. Participants express appreciation for the solution provided by a user named greg1313. The calculation involves substituting values into the function and summing the results over the specified range. The clarity of the solution and its effectiveness in addressing the problem are highlighted. Overall, the thread emphasizes the successful application of mathematical principles to achieve the desired sum.
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Let

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

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

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 $\dfrac{2017}{2}$.
 
greg1313 said:
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 !
 
greg1313 said:
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!
 
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