# The General Formula for the Difference of Two Nth Powers For all $a, b$ and $n \ge 2$: $ a^n - b^n = (a-b)\sum_{k=0}^{n-1}[a^{n-k-1}b^k] $ ## Proof $ \begin{align} (a-b)\sum_{k=0}^{n-1}[a^{n-k-1}b^k] &= \sum_{k=0}^{n-1}[a^{n-k}b^k] - \sum_{k=0}^{n-1}[a^{n-k-1}b^{k+1}] \\ &= a^n + \sum_{k=1}^{n-1}[a^{n-k}b^k] - \sum_{k=0}^{n-2}[a^{n-k-1}b^{k+1}] -b^n \end{align} $ Focus on the second sum. If we started with 1 instead of zero we would have to increase the element count by 1 and also we'd have to modify the expression inside the sum (to decrease $k$). $ \begin{align} \sum_{k=0}^{n-2}[a^{n-k-1}b^{k+1}] &= \sum_{k=1}^{n-1}[a^{n-(k-1)-1}b^{(k-1)+1}] \\ &= \sum_{k=1}^{n-1}[a^{n-k}b^k] \end{align} $ So the two sums in the statement above cancel out!