Today was a fruitful day for math. I watched more of the Essence of Calculus series, did a couple of pages of algebra review exercises, then started on the book The Joy of X by Steven Strogatz, which I bought earlier in the week. Near the beginning of that book, the author mentions that the $n$th square number is the sum of the first $n$ odd numbers. I stopped reading at that point and went to see if I could prove the claim. I was able to come up with two proofs, as well as a way of visualizing the problem geometrically that led me to a formula for the sum of the first $n$ even numbers. I also checked that the sum of the two formulas yielded the well known one for the sum of the first $n$ natural numbers, which it did.
Proposition:
$$\sum_{i=1}^{n}(2i-1)=n^2\text{ for all }n\in\mathbb{N}\text{.}$$
The proof by induction is easy…
Proof 1 (by induction):
When $n=1$, $\sum_{i=1}^{n}(2i-1)=\sum_{i=1}^{1}(2i-1)=2(1)-1=1=1^2$.
Assume that $\sum_{i=1}^{n-1}(2i-1)=(n-1)^2$.
It follows that $\sum_{i=1}^{n}(2i-1)=\sum_{i=1}^{n-1}(2i-1)+(2n-1)=(n-1)^2+2n-1=$ $(n^2-2n+1)+2n-1=n^2$.
Therefore, by induction, the proposition holds for all $n\in\mathbb{N}$.
Proof 2:
For any $n\in\mathbb{N}$, $\sum_{i=1}^{n}(2i-1)=[2n-1]+[2(n-1)-1]+[2(n-2)-1]+…+[2(1)-1]$.
Since there are $n$ terms in the sum, there are $n$ instances of $-1$.
Thus, $\sum_{i=1}^{n}(2i-1)=2[n+(n-1)+(n-2)+…+1]-n$.
Using the well known formula for the sum of the first $n$ natural numbers, it follows that $\sum_{i=1}^{n}(2i-1)=2[\frac{n(n+1)}{2}]-n=n(n+1)-n=n[(n+1)-1]=n^2$.
Thus, the proposition holds for all $n\in\mathbb{N}$.
Stay tuned tomorrow for a geometric visualization of the problem and the formula for the sum of the first $n$ even numbers.
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