# Induction Proof The sum of the first n odd squares

I am just learning about induction proofs. So far I am only familiar with induction equality proofs, and inequality proofs. Such as, for example, prove the sum of the powers of 2 = $2^{n+1} - 1$...

I am confused on the following proof: The sum of the first n odd squares is $\frac 43 n^3 - \frac 13n$

How do I start this proof? it looks like only one statement with nothing to compare it to. I was thinking maybe I would represent the sum of the first n odd squares as the formula $(2n - 1)^2$ and then set the proof up as $(2n - 1)^2 = \frac 43 n^3 - \frac 13n$

so it looks more like the form I am used to. Is this correct? Am I missing a small nuance of importance? Thanks for any and all help.

• Yes, I would be representing it as a summation notation containing (2i - 1)^2 – Conner May 18 '17 at 1:17
• I think $(2n+1)^2$ is better but not sure if it matters. – marshal craft May 18 '17 at 1:18

## 2 Answers

You need to take the sum of the first $n$ odd squares, e.g.if $n=3$, then you need to add $1+9+25=35$. And, that does indeed equal $\frac{4}{3}n^3-\frac{1}{3}n$ for $n=3$: $\frac{4}{3}3^3-\frac{1}{3}3=\frac{4}{3}27-1=36-1=35$

Now, to prove that this is true in general using induction:

Base: $n=0$: $\frac{4}{3}0^3-\frac{1}{3}0=0-0=0$ which is indeed the sum of the first $0$ odd squares. Check!

Step: Assume that for some $k$ the sum of the first $k$ odd squares is $\frac{4}{3}k^3-\frac{1}{3}k$. Now let's consider the sum of the first $k+1$ odd squares, which is of course the sum of the first $k$ odd squares plus the $k+1$-th odd square, which is $(2k+1)^2$. So, by the inductive hypothesis the sum is $\frac{4}{3}k^3-\frac{1}{3}k+(2k+1)^2$, and now you need to verify that this does indeed equal $\frac{4}{3}(k+1)^3-\frac{1}{3}(k+1)$. Let's see:

$$\frac{4}{3}k^3-\frac{1}{3}k+(2k+1)^2=$$

$$\frac{4k^3-k+3(4k^2+4k+1)}{3}=$$

$$\frac{4k^3+12k^2+12k-k+3}{3}=$$

$$\frac{4k^3+12k^2+12k+4-k+3-4}{3}=$$

$$\frac{4(k^3+3k^2+3k+1)-(k+1)}{3}=$$

$$\frac{4(k+1)^3}{3}-\frac{1}{3}(k+1)$$

• @Conner You're welcome! Please note that I had a small mistake in my earlier answer ... The $(k+1)$-th odd square is $(2k+1)^2$, and not $(2k-1)^2$, which is what I had first. – Bram28 May 18 '17 at 1:34
• Thank you. Just one more clarity question. Specifically when you are constructing your Induction Hypothesis, I was under the impression from earlier proofs I've done that we would have to insert K + 1 into (2k + 1)^2, making it (2(k + 1))^2 = (2k + 3)^2, and then we would add that. Is that not true in this case? – Conner May 18 '17 at 2:01
• @Conner The $k$ -th odd square is $(2k-1)^2$, and so the $(k+1)$-th odd square is $(2(k+1)-1)^2=(2k+1)^2$. So yes, you do fill in $k+1$ for $k$, but you have to choose the right formula! :) – Bram28 May 18 '17 at 2:09
• Ahh. Yes I see. I was very confused for a moment because I changed my entire representation of the summation of the odd squares to (2n+1)^2 instead, so I could account for the base n = 0. It didn't even cross my mind that you were just showing the example of if I had kept it (2n-1)^2. This has really helped my comprehension the induction proof process to me though. I appreciate you answering back. – Conner May 18 '17 at 2:13

I will give an alternative method to prove the claim.

It's a standard exercise in induction to show that \begin{align} \sum^{2N}_{n=1} n^2 = \frac{2N(2N+1)(4N+1)}{6}. \end{align} Next, observe \begin{align} \sum_{n \text{ even}}+\sum_{n\text{ odd}} n^2 =& \sum^N_{k=1} (2k)^2+\sum_{n \text{ odd}}^{2N}n^2\\ =& 4\sum^N_{k=1}k^2 + \sum_{n \text{ odd}} n^2\\ =&\ \frac{4N(N+1)(2N+1)}{6}+\sum_{n \text{ odd}} n^2. \end{align} Hence it follows \begin{align} \sum_{n \text{ odd}}^{2N} n^2 =&\ \frac{2N(2N+1)(4N+1)}{6}- \frac{4N(N+1)(2N+1)}{6}\\ =&\ \frac{4N^3-N}{3} \end{align}

Now, use induction to prove the above formula holds if you like.