If the sum $\frac{1 \times 2}{1^3} + \frac{2 \times 3}{1^3 +2^3} + \frac{3 \times 4}{1^3 + 2^3 + 3^3} + \frac{4 \times 5}{1^3 + 2^3 + 3^3 +4^3} + ... + $ up to 20 terms equals to $\frac{k}{21}$, then $k= ?$
I have found out that $\sum\limits^{n}_{i=1} i^3 = [\frac{n(n+1)}{2}]^2$, and that $k = 80$ based on adding up everything, but how do you find the general term for this series?
Based on $\sum\limits^{n}_{i=1} i^3 = [\frac{n(n+1)}{2}]^2$, each terms becomes $$\sum_n{n(n+1)\over [{n(n+1)\over 2}]^2}=\sum_n{4\over n(n+1)}$$which is a simple telescopic series.