The Sum of Raising the First \(n\) Integers to a Given Power
The closed-form expression for \(\sum_{i=1}^n i^m\) for a given \(m\in\mathbb{Z}^+\) can be derived — once we know the formulas for all lesser \(m\)’s — as follows:
First, observe that:
\begin{align*} n^{m+1} & = n^{m+1} - (n-1)^{m+1} \\ & + (n-1)^{m+1} - (n-2)^{m+1} \\ & + (n-2)^{m+1} - (n-3)^{m+1} \\ & + \cdots \\ & + 2^{m+1} - 1^{m+1} \\ & + 1^{m+1} - 0^{m+1} \\ & = \sum_{i=1}^n (i^{m+1} - (i-1)^{m+1}) \end{align*}Expand \(i^{m+1} - (i-1)^{m+1}\) for the \(m\) in question and use the linear transformation nature of summation to rewrite the right-hand side of \(n^{m+1} = \sum_{i=1}^n (i^{m+1} - (i-1)^{m+1})\) as a sum of summations of \(i^k\)’s.
Isolate the \(\sum_{i=1}^n i^m\) term of the equation.
Substitute the closed-form expressions for the remaining \(\sum_{i=1}^n i^k\) terms.
For example, once we know that \(\sum_{i=1}^n i = \frac{n(n+1)}{2}\) and \(\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}\), we can derive \(\sum_{i=1}^n i^3\) as follows:
\begin{align*}
n^4 & = \sum_{i=1}^n (i^4 - (i-1)^4) \\
& = \sum_{i=1}^n (4i^3 - 6i^2 + 4i - 1) \\
& = 4\sum_{i=1}^n i^3 - 6\sum_{i=1}^n i^2 + 4\sum_{i=1}^n i - n \\
\sum_{i=1}^n i^3
& = \frac{1}{4} (n^4 + 6\sum_{i=1}^n i^2 - 4\sum_{i=1}^n i + n) \\
& = \frac{1}{4} (n^4 + 6\times\frac{n(n+1)(2n+1)}{6} - 4\times\frac{n(n+1)}{2} + n)\\
& = \frac{1}{4} (n^4 + 2n^3 + n^2)
\end{align*}