How to evaluate $\sum_{k=1}^{n} \frac{f^3_k}{3(f^2_k-f_k)+1}, ~~f_k=k/n$ Very often the summations at pre-degree level are done by using differencing-telescoping and by some  symmetry properties. The following summation is fabricated keeping one thing in mind, I may tell it later. $$\sum_{k=1}^{n} \frac{f^3_k}{3(f^2_k-f_k)+1},~~ f_k=k/n.$$
The question is: How will you evaluate this sum?
 A: let $$S= \sum_{k=0}^{n} \frac{f_k^3}{3(f_k^2-f_k)+1}=\sum_{k=1}^{n} \frac{f_k^3}{3(f_k^2-f_k)+1}$$ writing sum in reverse order $$S=\sum_{k=0}^n\frac{f_{n-k}^3}{3(f_{n-k}^2-f_{n-k})+1}$$ now using the beatiful property $f_k=1-f_{n-k}$ $$S=\sum_{k=0}^n \frac{{(1-f_k)}^3}{3(f_k^2-f_k)+1}$$ Adding $$2S=\sum_{k=0}^n 1=n$$ $$S=(n+1)/2$$
A: I ended up with a similar solution to that of Albus Dumbledore but I wonder why Ablus didn't make it more obvious.
Notice that the numerator invites for a third order binomial manipulation:
$$3(f^2_k-f_k)+1 = f_k^3 + (1-f_k)^3 = f_k^3 + f_{n-k}^3$$
Therefore
$$S = \sum_{k=0}^{n} \frac{f^3_k}{3(f^2_k-f_k)+1} = \sum_{k=0}^{n} \frac{f^3_k}{f_k^3 + f_{n-k}^3} \\
= \frac 12 \sum_{k=0}^{n} \left(\frac{f^3_k}{f_k^3 + f_{n-k}^3} + \frac{f^3_{n-k}}{f_k^3 + f_{n-k}^3} \right)= \frac 12 (n+1).\blacksquare$$
A: There a many ways to do it, I presume.
$$S_n=\sum_{k=1}^n \frac{k^3}{3 k^2 n-3  n^2k+n^3}=\sum_{k=1}^n \frac{k^3}{3(k-a)(k-b) }$$ where
$$a=\frac{n}{6} \left(3-i \sqrt{3}\right)\qquad \text{and} \qquad b=\frac{n}{6} \left(3+i \sqrt{3}\right)$$
$$S_n=\sum_{k=1}^n \Bigg[\frac{a^3}{3 (a-b) (k-a)}-\frac{b^3}{3 (a-b) (k-b)}+\frac{a+b}{3}+\frac{k}{3}\Bigg]$$
$$S_n=\frac{2 a^3 \left(H_{n-a}-H_{-a}\right)+n (a-b) (2 a+2 b+n+1)+2 b^3
   \left(H_{-b}-H_{n-b}\right)}{6 (a-b)}$$
$$S_n=\frac{1}{18} \left(9 n+9+\frac{4 \pi  n \sin (\pi  n)}{\cos (\pi  n)-\cosh
   \left(\frac{\pi  n}{\sqrt{3}}\right)}\right)=\frac{n+1}2$$
