How to show $\sum_{k=1}^{80} {1 \over \frac{k}{81} - \frac{1}{2} - \frac{\iota}{2}} + {1 \over \frac{k}{81} - \frac{1}{2} +\frac{\iota}{2}} = 0$? I came across this sum while doing the homework my teacher gave us on series. It was originally $$a_k = \frac{k}{81}, S = \sum_{k=1}^{80} {a_k^2 \over 1+2a_k^2-2a_k}$$
I then decomposed it into partial fractions and ended up with
$40+ \sum_{k=1}^{80} {1 \over \frac{k}{81} - \frac{1}{2} - \frac{\iota}{2}} + {1 \over \frac{k}{81} - \frac{1}{2} +\frac{\iota}{2}}$. I know the answer is 40 from WolframAlpha, however I don't know how to go about showing that the sum with the complex numbers is 0.
How should I proceed?
 A: There is no need to use complex numbers. Here the keyword is symmetry: note that $a_{81-k}=1-a_k$ and therefore
\begin{align}
2S&=\sum_{k=1}^{80} \frac{a_k^2}{1-2a_k(1-a_k)}+\sum_{k=1}^{80} \frac{a_{81-k}^2}{1-2a_{81-k}(1-a_{81-k})}\\
&=\sum_{k=1}^{80} \frac{a_k^2}{1-2a_k(1-a_k)}+\sum_{k=1}^{80} \frac{(1-a_k)^2 }{1-2(1-a_k)a_k}\\
&=\sum_{k=1}^{80} \underbrace{\frac{a_k^2+(1-a_k)^2}{1-2a_k(1-a_k)}}_{=1}=80
\end{align}
which implies that $S=40$.
A: Notice that :
$$S = \sum_{k = 1}^{80} \dfrac{k^2}{81^2 + 2 k^2 - 2 \times 81 \times k} = \sum_{k = 1}^{80} \dfrac{k^2}{(81 - k)^2 + k^2}$$
and :
$$\sum_{k = 1}^{80} \dfrac{k^2}{(81 - k)^2 + k^2} = \sum_{k = 1}^{80} \dfrac{(81 - k)^2}{k^2 + (81 - k)^2}$$
which allows us to deduce that :
$$2 S = \sum_{k = 1}^{80} \dfrac{k^2}{(81 - k)^2 + k^2} + \sum_{k = 1}^{80} \dfrac{(81 - k)^2}{k^2 + (81 - k)^2} = \sum_{k = 1}^{80} \dfrac{k^2 + (81 - k)^2}{k^2 + (81 - k)^2} = \sum_{k = 1}^{80} 1 = 80$$
Hence :
$$S = 40$$
A: Just some observations that can serve as "food for thought" to solve these problems:

*

*Remember the proof of little Gauss that $1+2+...+(n-1)+n=n(n+1)/2$. The idea was to combine terms in the sum so that the sum is constant. In this case we can combine $(1,n),(2,n-1),(3,n-2)$ with constant sum $n+1$. Suppose $n$ is even as a start. Since we have $n/2$ such terms, the answer follows ;


*If we have a new problem/sum how can we get an intuition on how to combine terms (supposing this technique makes sense to apply)? Either we are in a good day and see the solution (like the answers already posted), or we can start evaluating the sum for small $n$ by hand and see if we see a pattern, that we can conjecture and then verify.
In this case for example we could try to generalize:
$80 \rightarrow 2n, a_k \rightarrow \frac{k}{2n+1}$
and if we call:
$\beta_k=\frac{a_k^2}{1+2a_k^2-2a_k}$
it is not difficult to conjecture, looking at cases $n=1,2$, that:
$\beta_{2n+1-k}+\beta_{k}=1$,
independently of $k$. Once we verify this identity (some calculations), we know that:
$\sum_{k=1}^{2n} \beta_k=n$,
which is the correct answer for every $n$.
