Good idea to find below sum this is the question.
Purpose is to find sum of $$f\left(\frac{1}{14}\right)+f\left(\frac{2}{14}\right)+f\left(\frac{3}{14}\right)+...+f\left(\frac{13}{14}\right)$$for function $$f(x)=\frac{4^x}{4^x+2}$$Is there a nice method to find $f\left(\frac{1}{14}\right)+f\left(\frac{2}{14}\right)+f\left(\frac{3}{14}\right)+...+f\left(\frac{13}{14}\right)$ ?
 A: Yup, a nice thing to notice is the value of
$$
\begin{align}
f(x)+f(1-x)
&=\frac{4^x}{4^x+2}+\frac{4}{4+2\cdot4^x}\\
&=\frac{4^x}{4^x+2}+\frac{2}{4^x+2}\\[3pt]
&=\cdots\,?
\end{align}
$$
A: As remarked by @Nikunj, the function defined by $f(x)=\dfrac{4^x}{4^x+2}$ has a center of symmetry at $(1/2, f(1/2))$. In particular, for derivatives:
$$\forall n \ \ f^{(n)}(0)=f^{(n)}(1) \ \ \ \ \ (1)$$
We are going to apply the Euler Maclaurin formula, or more exactly one of its versions, which connects a sum and an integral in the following way:
$$\sum_{k=0}^m f(a+kh)=\frac{1}{h}\int_a^b f(x)dx+\frac{1}{2}(f(a)+f(b))+\sum_{k=1}^{\infty} \left(\frac{h}{2}\right)^{2k-1}b_{2k}\left(f^{(2k-1)}(b)-f^{(2k-1)}(a)\right)$$
where line segment $[a,b]$ has been divided into $n$ parts of size $h=\frac{b-a}{n}$, with "universal" coefficients $b_k$ linked to Bernoulli numbers.
It gives, in our case, with $a=0$, $b=1$, $n=14$, $h=\frac{1}{14}$, by using relationship (1):
$$\sum_{k=0}^{14} f\left(0+\frac{k}{14}\right)=14\int_0^1 f(x)dx+\frac{1}{2}\left(\frac{1}{3}+\frac{2}{3}\right)+0$$
The sum we have to compute is the previous one without the first and last term: $\frac{1}{3}$ and $\frac{2}{3}$. Thus:
$$\sum_{k=1}^{13}f\left(\frac{k}{14}\right)=14\int_0^1 f(x)dx-\frac{1}{2}=14 \times \frac{1}{2}-\frac{1}{2}=\frac{13}{2}$$
which is the numerical answer.
Explanation for the last inequality: the integral of $f$ is easily computed, with value $\frac{1}{2}$, due to a "nice" primitive function $\dfrac{1}{\ln{4}}\ln(2+4^x)$.
See p. 806 of "Handbook of Mathematical Functions", M. Abramowitz and I. Segun, Dover, 1972.
See also formula (7) page 6 of http://www.hep.caltech.edu/~phys199/lectures/lect5_6_ems.pdf
Remarks about Maclaurin formula:
a) A more appropriate way to write it is with a finite sum and a remainder under an integral form.
b) This formula extends the trapezoidal rule of integration.
Edit 1: In fact, function $f$ is connected to $\tanh$ :
$$f(x)=\dfrac{4^x}{4^x+2}=\dfrac{1}{1+2 \times 4^{-x}}=\dfrac{1}{1+e^{\ln 2 - x \ln 4}}=\dfrac{1}{1+e^{\ln 2(1 - 2x)}}$$
$$f(x)=\frac12(1 + \tanh(a(x - \tfrac12))) \ \text{with} \ a=\ln 2 \tag{1}$$
(due to formula : $\frac12 \tanh x = \frac12 - \frac{1}{1+e^{2x}}$)
Edit 2: Here is a graphical representation of function $f$ obtained by a succession of affine transformations from the graphical representation of function $\tanh$) :

