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?

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 $$2S=\sum_{k=0}^n 1=n$$ $$S=(n+1)/2$$
