$$1 = 1$$ $$1 + 2 + 1= 4 = 2^2$$ $$1 + 2 + 3 + 2 + 1 = 9 = 3^2$$ $$1 + 2 + 3 + 4 + 3 + 2 + 1 = 16 = 4^2$$

See a pattern? It looks like the following formula holds for integers $n>0$:

$$1+2+3+\cdots+ (n-1) + n + (n-1) + \cdots + 1 = n^2$$

There is an intuitive geometric explanation for this formula. Consider arranging an $n$ by $n$ square lattice of beads. There is a total of $n^2$ beads in this lattice. We can imagine many different ways to walk through the lattice and count them; all such ways will lead to the answer $n^2$. One way to walk through the lattice is to start at a corner and count along the diagonals. If we walk through the entire square lattice of beads like this, for $n=4$, we would count $1+2+3+4+3+2+1$– where the $4$ contribution comes from the main diagonal and the $1$ contributions come from the two opposing corners– and get $n^2=16$ since this is the total number of beads.

Applied Mathematician Steven Strogatz posted a photo on Twitter that illustrates this geometric interpretation of the formula $1 + 2 + 3 + 4 + 3 + 2 + 1 = 16 = 4^2$. Each type of bead classifies a diagonal on the square lattice of beads.

Proof by beads: 1+2+3+4+3+2+1 = 4^2. Same idea with square of side n shows 1+2+...+ n + ... +2+1 =n^2. pic.twitter.com/uxKGxwQGAW

— Steven Strogatz (@stevenstrogatz) February 22, 2015

The more rigorous way to prove

$$S(n):=1+2+3+\cdots+ (n-1) + n + (n-1) + \cdots + 1 = n^2$$

is by Mathematical induction.

The base case is $n=1$. Of course, $S(1)=1$, i.e., $1=1$ is true.

The inductive step is to assume that $S(n)=n^2$ holds and then show that this implies $S(n+1)=(n+1)^2$.

Somehow, we need to construct $(n+1)^2$ on the right-hand side of $S(n)=n^2$. Noting that $(n+1)^2=n^2+2n+1$, we can add $2n+1$ to both sides

$$S(n) + [2n+1]= n^2 + [2n+1] = (n+1)^2.$$

The left-hand side can be rewritten by noting that $2n+1=n+(n+1)$:

$$S(n)+ [n+ (n+1)] = 1+2+\cdots+ (n-1) + n + (n+1) +n + (n-1) + \cdots + 1.$$

Can you recognize this as $S(n+1)$? We have just proven that $S(n+1)=(n+1)^2$ starting with $S(n)=n^2$. That is, we have proven:

$$1+2+\cdots+ (n-1) + n + (n+1) +n + (n-1) + \cdots + 1=(n+1)^2,$$

completing the proof by induction.

The geometric interpretation is much nicer right?

comments powered by Disqus