How to determine equation for $\sum_{k=1}^n k^3$ How do you find an algebraic formula for $\sum_{k=1}^n k^3$? I am able to find one for $\sum_{k=1}^n k^2$, but not $k^3$. Any hints would be appreciated.

One classic method is
$$\begin{align} \sum_{k=1}^{n} k^4 &= \left( \sum_{k=1}^n (k+1)^4 \right)  + 1 - (n+1)^4
\\&= \left( \sum_{k=1}^n k^4 + 4 k^3 + 6 k^2 + 4 k + 1 \right)  + 1 - (n+1)^4
\\&= \left( \sum_{k=1}^n k^4 \right)
+ 4 \left( \sum_{k=1}^n k^3 \right)
+ 6 \left( \sum_{k=1}^n k^2 \right)
+ 4 \left( \sum_{k=1}^n k^1 \right)
+ \left( \sum_{k=1}^n 1 \right)
+ 1 - (n+1)^4
\end{align}$$
The sum of fourth powers cancel out, and you can solve the equation for the sum of curves.

Another approach is to guess that the sum of cubes should be a fourth degree polynomial, then solve for the coefficients of the polynomial using the equations


*

*$f(0) = 0$

*$f(1) = 1$

*$f(2) = 1 + 2^3$

*$f(3) = 1 + 2^3 + 3^3$

*$f(4) = 1 + 2^3 + 3^3 + 4^3$


Five equations in five unknowns. You can cut this down to 2 unknowns if you recognize some easy patterns: the leading coefficient for the sum of $n$-th powers is always $1/n$ and the next is always $1/2$, and the constant term is always 0.
