Sum of $\lfloor k^{1/3} \rfloor$ I am faced with the following sum:
$$\sum_{k=0}^m \lfloor k^{1/3} \rfloor$$
Where $m$ is a positive integer. I have determined a formula for the last couple of terms such that $\lfloor n^{1/3} \rfloor^3 = \lfloor m^{1/3} \rfloor^3$. For example if the sum is from 0 to 11 I can find the sum of the terms 8 through 11. I am stuck on what formula can be applied to find the sum of the terms before the final stretch however. 

You have
$$
\begin{align*}
\sum_{k=0}^m \lfloor k^{1/3} \rfloor &= \sum_{i=0}^{\lfloor m^{1/3} \rfloor} i |\{0 \leq k \leq m : \lfloor k^{1/3} \rfloor = i\}| \\ &=
\sum_{i=0}^{\lfloor m^{1/3} \rfloor} i |\{0 \leq k \leq m : i^3 \leq k < (i+1)^3\}| \\ &=
\sum_{i=0}^{\lfloor m^{1/3} \rfloor-1} i((i+1)^3-i^3) + \lfloor m^{1/3} \rfloor (m-\lfloor m^{1/3} \rfloor^3+1) \\ &=
\sum_{i=0}^{\lfloor m^{1/3} \rfloor-1} (3i^3+3i^2+i) + \lfloor m^{1/3} \rfloor (m-\lfloor m^{1/3} \rfloor^3+1) \\ &=
\frac{1}{4}(\lfloor m^{1/3} \rfloor-1)\lfloor m^{1/3} \rfloor^2(3\lfloor m^{1/3} \rfloor+1) + \lfloor m^{1/3} \rfloor (m-\lfloor m^{1/3} \rfloor^3+1).
\end{align*}
$$
