What is a good way to prove that the sum of the first n odd natural numbers is n²?

Question

We can try to prove the statement by mathematical Induction technique or by using the following method: $$S=1+3+5...+(2n-1)$$ $$S=(2n-1)+(2n-3)+(2n-5)...+1$$ $$2S=2n+2n+2n...+2n$$ $$2S=2n^2$$ This method also includes mathematical Induction (as people pointed out). My question is do these two methods of proving works independently, since, I have doubt that they may be integrated or combined for a more formal (or better) proof.

Answer 1

Both are valid proofs, combining them does not result in a 'more formal' proof. But to answer the question in the title, the best way of proving this fact is a picture:

(picture taken from https://www.math.upenn.edu/~deturck/probsolv/LP1ans.html)

Answer 2

What exactly $S_n=1+3+\cdots+(2n-1)$ means?

Well, technically there's a recursive definition hidden in the "$\cdots$" symbol:

$$S_1=1$$ $$S_n=S_{n-1} + 2n-1$$

And recursion is equivalent to induction.

Now secondly, you need to know that $a_1+\cdots +a_n=a_n+\cdots+a_1$. While this follows from the commutativity of addition, formally you would need induction to prove this. Note that $S'_n=a_n+\cdots + a_1$ has a different (even though similar) recursive definition:

$$S'_1=1$$ $$S'_n=2n-1+S'_{n-1}$$

Finally, you need to prove that $a_1+\cdots+a_n+b_1+\cdots+b_n=(a_1+b_1)+\cdots+(a_n+b_n)$ and $a+\cdots +a$ added $n$-times is equl to $n\cdot a$ which both need induction as well.

And so "yes", this is a valid proof, but "no", this is not a proof without induction. Your proof heavily depends on the induction. The induction is just well hidden there.

Answer 3

What you did is correct, you can also show it by induction, which is different from your method above.

Induction start:

For $n=1$ we have $1=1^2$. For $n=2$ we have $1+3=4$ and $2^2=4$, which is again true.

Induction hypothesis (IH):

Assume it holds for the $n$th odd natural number $(2n-1)$, that is $1+3+5+....+(2n-1) =n^2.$

Induction step:

Using IH we want to show that it also holds for the first $(n+1)$ odd natural numbers. Indeed we have $$1+3+5+....+(2n-1)+\color{red}{2n+1}\underbrace{=}_{\text{by $IH$}}n^2+2n+1$$ $$=(n+1)^2$$

Thus if we assume that it is true for any arbitrary odd natural number $n$, it must be true for the next odd natural number $n+1$. Since it is true for the first natural number, applying the result infinitely many times, we conclude that it is true for all natural numbers.

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Another way.

Since for the sum of $n$ natural numbers we have $$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$$

Which can be shown by induction or your method above for example. Then for the sum of of the first $n$ odd natural numbers we have $$\sum_{k=1}^{n}(2k-1)$$ $$=2\sum_{k=1}^{n}k-\sum_{k=1}^{n}1=2\frac{n(n+1)}{2}-n$$ $$=n(n+1)-n=n^2+n-n=n^2.$$