Sum of the First $N$ Natural Numbers

11 Jun 2016

Lately I’ve been thinking about the content of my high school mathematics courses. In my college algebra and trigonmetry class we were introduced to proofs by induction–of all things–somewhere towards the end of my junior year. An important proof technique, for sure. However, the combinatorial expressions they were applied to, like the sum of first $N$ natural numbers, seemed miraculous. Proof by induction is fairly straight forward. But how does one even guess at a closed formula for such expressions? I remember the teacher saying something to the effect “you need to be smart”. I suppose so, but we can actually construct the closed form of the sum of the first $N$ natural numbers.

Let

\[f(N) = 1 + 2 + \cdots + (N-1) + N. \\\]

Rearranging the sum doesn’t change the value of the expression so $ 2 \times f(N)$ can also be written

\[\begin{array}[c]{} & 1 & + & 2 & + & \cdots & + & N-1 & + & N \\ \Large{+} & N & + & N-1 & + & \cdots & + & 2 & + & 1 \\ \hline & N+1 & + & N+1 & + & \cdots & + & N+1 & + & N+1 \end{array}\]

Above $f(N)$ is added to itself, however the second instance of $f(N)$ is written as a sum in reverse order. The $N$ intermediate (or column) sums are all $N+1$. Therefore,

\[\begin{gather*} 2 \times f(N) = N(N+1) \quad \Longrightarrow \quad f(N) = \binom{N+1}{2}. \end{gather*}\]