lfdahl said:Let
\[f(x) = \frac{a^{2x}}{a^{2x}+a}, \;\;\; a \in \Bbb{N}.\]Find the sum:\[ \sum_{j=0}^{2016}f \left ( \frac{j}{2016} \right )\]
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!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 $\dfrac{2017}{2}$.
The formula for calculating the sum of f(x) from 0 to 2016 is ∑f(x) = f(0) + f(1) + f(2) + ... + f(2016), where f(x) is the function and x is the variable.
To determine the value of f(x) for each term, simply plug in the value of x into the function. For example, if f(x) = 2x + 3, then f(0) = 3, f(1) = 5, f(2) = 7, and so on.
Yes, the sum of f(x) from 0 to 2016 can be calculated using a calculator. Simply enter the formula ∑f(x) = f(0) + f(1) + f(2) + ... + f(2016) into your calculator and replace each term with its corresponding value.
Calculating the sum of f(x) from 0 to 2016 can be useful in various mathematical contexts, such as finding the area under a curve or determining the average or total value of a function over a certain interval.
Yes, there is a shortcut method called the summation formula or sigma notation, which allows you to write the sum of a function over a certain interval in a more concise and efficient way. It is written as ∑f(x) = ∑(from x=0 to 2016) f(x).