I have been on this problem and my professor has given me anything useful to work with. $$\forall n\geq 1 :\quad \sum_{i=1}^{2n} i^3 = n^2(2n+1)^2$$
This is all the work I have currently on this problem.
Base case n = 1 $$\sum_{i=1}^{2n} i^3 = 1^3+2^3 = 9$$ $$n^2(2n+1)^2 = (1)^2(2(1)+1)^2 = 9$$ Suppose $$\sum_{i=1}^{2k} i^3 = k^2(2k+1)^2$$ Need to show $$\sum_{i=1}^{2k+2} i^3 = (k+1)^2(2(k+1)+1)^2$$ observe $$\sum_{i=1}^{2k+2} i^3 =\sum_{i=1}^{2k} i^3+(2k+1)^3+(2k+2)^3$$ Inductive Hypothesis says $$=k^2(2k+1)^2+(2k+1)^3+(2k+2)^3$$ $$=(2k+1)^2(k^2+2k+1)+(2k+2)^3$$ $$=(2k+1)^2((k+1)^2)+2^3(k+1)^3$$ $$=(k+1)^2((2k+1)^2+(k+1)2^3)$$ $$=(k+1^2(2k+1)^2+(8k+8))$$ $$=(k+1)^2(4k^2+4k+8k+9)$$ $$=(k+1)^2(4k^2+12k+9)$$ $$=(k+1)^2(2k+3)^2$$ Been stuck here for a while and have no idea where to go...
