Special Sums: Sum of the First $n$ Natural Numbers, Their Squares, and Their Cubes

Three particular sums recur so often — both later in this syllabus and in the two exercises above — that they are worth deriving once and for all: the sum of the first $n$ natural numbers, of their squares, and of their cubes.

Sum of the first $n$ natural numbers, $\sum_{k=1}^n k$. The natural numbers $1,2,3,\ldots,n$ are themselves an A.P. with first term $a=1$, common difference $d=1$, and last term $l=n$. Applying the A.P. sum formula of Section 2 directly,

(This is exactly the "reverse and add" computation Gauss is said to have used: $1+2+\cdots+n$ added to its own reverse $n+(n-1)+\cdots+1$ gives $n$ copies of $(n+1)$, so the sum is half of $n(n+1)$.)

Sum of squares, $\sum_{k=1}^n k^2$ — derivation by telescoping. Start from the algebraic identity

which holds for every integer $k$ (expand $(k+1)^3=k^3+3k^2+3k+1$ and subtract $k^3$). Summing both sides for $k=1,2,\ldots,n$, the left side telescopes — every intermediate value cancels, leaving only the very first and very last:

The right side sums to $3\sum k^2 + 3\sum k + n$ (using $\sum_{k=1}^n 1=n$). So

Solving for $\sum k^2$ and simplifying the right-hand side (expand $(n+1)^3=n^3+3n^2+3n+1$, collect like powers of $n$, and factor) gives the standard closed form:

Sum of cubes, $\sum_{k=1}^n k^3$ — derivation by telescoping (same method, one power higher). Starting instead from $(k+1)^4-k^4 = 4k^3+6k^2+4k+1$ and summing $k=1$ to $n$, the left side again telescopes to $(n+1)^4-1$, giving an equation in $\sum k^3$ once $\sum k^2$ and $\sum k$ (both already known above) are substituted in. Carrying out the same style of algebraic simplification as for the squares yields the remarkably clean closed form

$\boxed{\sum_{k=1}^n k^3 = \left[\frac{n(n+1)}{2}\right]^2 = \left(\sum_{k=1}^n k\right)^2}.$ …