# Sum of Cubes of First N Integers $\sum_{n=1}^b n^3 = \left(\frac{b(b+1)}{2}\right)^2=\left(\sum_{n=1}^b n\right)^2$ ^x3uvpu ## Proof by Binomial Expansion Start with binomial expansion of $(n-1)^4$ [^1]: $ \begin{gather} (n-1)^4=n^4-4n^3+6n^2-4n+1 \\ n^4 - (n-1)^4=4n^3-6n^2+4n-1 \end{gather} $ [^1]: Per [[Binomial Theorem]] We can take the sum of both sides from $n=1$ to $b$ to get: $\sum_{n=1}^b\left[n^4 - (n-1)^4\right] = \sum_{n=1}^b\left[4n^3-6n^2+4n-1\right]$ Lets look at the two sides one by one. The left sum can be expanded like: $ \begin{gather} \sum_{n=1}^b\left[n^4 - (n-1)^4\right] = \\ 1 + \\ 2^4-1+ \\ 3^4-2^4+ \\ 4^4 - 3^4 \\ +\ldots+ \\ b^4-(b-1)^4 \\ =b^4 \end{gather} $ It is said that this series telescopes: all terms cancel out until the last $b^4$. The right sum can be simplified as: $\sum_{n=1}^b\left[4n^3-6n^2+4n-1\right] = 4 \sum_{n=1}^b \left[n^3\right] - 6 \sum_{n=1}^b\left[n^2\right] + 4\sum_{n=1}^b [n] - b$ When we put the left and right sides together we get: $ \begin{gather} b^4 = 4 \sum_{n=1}^b \left[n^3\right] - 6 \sum_{n=1}^b\left[n^2\right] + 4\sum_{n=1}^b [n] - b \\ 4 \sum_{n=1}^b \left[n^3\right] = b^4 + 6 \sum_{n=1}^b\left[n^2\right] - 4\sum_{n=1}^b [n] + b \end{gather} $ [[The Gauss Trick]] tells us the value of $\sum_{n=1}^b n$, and [[Sum of Squares of First N Integers]] tells us the value of $\sum_{n=1}^b[n^2]$, so we can simplify: $ \begin{align} 4 \sum_{n=1}^b \left[n^3\right] &= b^4 + 6 \sum_{n=1}^b\left[n^2\right] - 4\sum_{n=1}^b [n] + b \\ &= b^4 + 6 \frac {b(b+1)(2b+1)} {6} - 4\frac {b(b+1)} {2} + b \\ &= b^4 + b(b+1)(2b+1) - 2b(b+1) + b \\ &= b\left( b^3 + 2b^2+b+2b+1-2b-2+1\right) \\ &=b\left( b^3+2b^2+b \right) \\ &=b^2\left(b^2+2b+1 \right) \\ 4 \sum_{n=1}^b \left[n^3\right] &= b^2(b+1)^2 \\ \sum_{n=1}^b \left[n^3\right] &= \frac {b^2(b+1)^2} {4} = \left( \frac {b(b+1)} {2} \right)^2 \end{align} $