Inductive step of $2^2 + 4^2 + \cdots + (2n)^2$

Question

I am trying to prove that $\sum_{k=1}^n (2k)^2 = \frac{2n(n+1)(2n+1)}{3}$. I managed to do the base case but I am stuck at the inductive step. Am I doing it wrong?

This is what I need to prove:

$$2^2 + 4^2 + \cdots + (2n)^2 = \frac{2n(n+1)(2n+1)}{3}$$

Here is the working I did. I feel that I did something wrong but I am not sure which portion.

Show $k+1$ is true: $$2^2 + 4^2 + \cdots + (2k)^2 + (2(k+1))^2 = \frac{2(k+1)((k+1)+1)(2(k+1)+1)}{3}$$ $$\frac{2k(k+1)(2k+1)}{3} + \frac{3(2(k+1))^2}{3} = \frac{2(k+1)((k+1)+1)(2(k+1)+1)}{3}$$

Answer 1

Your proof is ill-written, since as written it assumes the conclusion. It would be a good idea to explicitly state your induction hypothesis first:

---

Induction Hypothesis: For $n \ge 1$, assume the following is true: $$\sum_{k=1}^n (2k)^2 = \frac{2n(n+1)(2n+1)}{3}$$

---

Then, on the assumption the above is true, you want to show that the subsequent case, the $n+1$ case, also holds. Bear in mind the difference between the following two starting points:

---

Induction Step: $$\sum_{k=1}^{n+1} (2k)^2 = \frac{2(n+1)(n+2)(2n+3)}{3}$$

---

Induction Step: I want to show the following holds: $$\sum_{k=1}^{n+1} (2k)^2 = \frac{2(n+1)(n+2)(2n+3)}{3}$$

---

The only difference between these two is that line of text in the latter. The reason this is important is that it makes clear that you don't know for sure that expression holds. Students often start with the first example instead, and manipulate both sides of the equality to show it's true. While this has some potential validity if done carefully and can be enlightening as to how the proof should go, it ultimately always seems to go down the rabbit hole of "assume it's true, show it leads to a true expression," obfuscating various details and nuances.

The proper thing to do would be, instead, something like this... (To avoid giving you the answer to your problem, and for simplicity, I'll refer to a more basic induction: $$\sum_{k=1}^n k = \frac{n(n+1)}{2}$$

---

Base Case: Take $n=1$. Then $$\sum_{k=1}^1 k = 1 = \frac{1(1+1)}{2}$$ validating our base case.

Induction Hypothesis: For $n \ge 1$, assume the following is true: $$\sum_{k=1}^n k = \frac{n(n+1)}{2}$$ Induction Step: We seek to verify the $n+1$ case. Note that $$\begin{align*} \sum_{k=1}^{n+1} k &= (n+1) + \sum_{k=1}^n k \\ &= n+1 + \frac{n(n+1)}{2} \\ &= \frac{2(n+1)+n(n+1)}{2} \\ &= \frac{(n+1)(n+2)}{2} \end{align*}$$ completing our induction and ensuring the expression holds $\forall n \in \Bbb Z^+$.

---

(As a student, or for less trivial inductions, you'd also be expected to justify each of the equalities in the induction step.)

Notice a few things here. First, we actively utilize our induction hypothesis, by manipulating the summation from the $n+1$ case into something clearly involving the $n$ case, which is something we assume to hold. Also notice how it flows: we start with the summation, and manipulate it until the desired quantity appears. We don't at any point bring in that specific expression, we just manipulate the summation until we can invoke our hypothesis, and then simplify to get the desired expression to appear. This is more rigorous, logically-speaking; it's also easier to write than salvaging the common student fallacies I mentioned earlier, and it's altogether clearer and flows better.

Thus, this is what I would suggest doing in your case: notice that

$$\sum_{k=1}^{n+1} (2k)^2 = (2(n+1))^2 + \sum_{k=1}^n (2k)^2$$

and, from there, invoke your induction hypothesis, manipulating the result to get your desired result in subsequent equalities.

Answer 2

$\textbf{Solution:}$ For any integer $n \geq 1$, let $S(n)$ denote the statements $$S(n): \sum_{i=1}^{n}(2i)^{2}=\frac{2n(n+1)(2n+1)}{3}$$

$\textbf{Base step}(n=1):$ $S(1)$ says $\sum_{i=1}^{1}(2i)^{2}=4$, and this is true because $$\sum_{i=1}^{1}(2i)^{2}=(2(1))^{2}=4$$

$\textbf{Inductive step}$ $ S(k) \implies S(k+1):$ Fix some $k \geq 1$. Assume that $$\sum_{i=1}^{k}(2i)^{2}=\frac{2k(k+1)(2k+1)}{3}$$ holds. To be proved is that $$S(k+1): \sum_{i=1}^{k+1}(2i)^{2}=\frac{2(k+1)((k+1)+1)(2(k+1)+1)}{3}$$follows. Beginning with the left side of $S(k+1)$, we have

\begin{align} \sum_{i=1}^{k+1}(2i)^{2}&=&\left[\sum_{i=1}^{k}(2i)^{2} \right]+(2(k+1))^{2} \\ &=&\frac{2k(k+1)(2k+1)}{3}+4(k+1)^{2}\\ &=&\frac{2k(k+1)(2k+1)+12(k+1)^{2}}{3}\\ &=&\frac{2(k+1)\left[k(2k+1)+6(k+1)\right]}{3}\\ &=&\frac{2(k+1)(2k^{2}+7k+6)}{3}\\ &=&\frac{2(k+1)(k+2)(2k+3)}{3}\\ &=&\frac{2(k+1)((k+1)+1)(2(k+1)+1)}{3} \end{align}

on arrive at the right side of $S(k+1)$, thereby showing $S(k+1)$ is also true, completing the inductive step.

$\textbf{Conclusion:}$ By mathematical induction, it is proved that for all $n \geq 1$, the statement $S(n)$ is true.$\Box$