A couple of thoughts.
Basis step
[math]n = 0 \implies \sum_{i=0}^n (2i + 1)^2 = (2 * 0 + 1)^2 = 1^2 = 1.\\
\text {And } n = 0 \implies \dfrac{(n + 1)(2 * n + 1)(2n + 3)}{3} =\\
\dfrac{(0 + 1)(2 * 0 + 1) * (2 * 0 + 3)}{3} = \dfrac{1 * 1 * 3}{3} = 1.\\
\therefore \ n = 0 \implies \sum_{i=0}^n (2i + 1)^2 = \dfrac{(n + 1)(2 * n + 1)(2n + 3)}{3}.
[/math]
Do you see that the problem talks about non-negative integers and the first possible example is [imath](2 * 0 + 1)^2[/imath]? This is no big deal in terms of your overall thought process, but a lot of math is about being very careful about what you are saying.
OK. I now think you have a bit of subtle confusion that I see in many students. To dispel that confusuion, I use what I call the intuition step. (Do not put this step into your work; it is not mathematically necessary, and many teachers find it annoying.) It helps students understand what we are doing and WHY it is justified in a common sense way.
[math]\text {The basis step has proved that there exists}\\
\text {AT LEAST ONE non-negative integer n for which it is true that}\\
\sum_{i=0}^n (2i + 1)^2 = \dfrac{(n + 1)(2n + 1)(2n + 3)}{3}.\\
\text {Let k be an ARBITRARY one of those non-negative integers.}
[/math]
I hate it that we talk about an induction “hypothesis.” We have proved in the basis step that at least one number does have the properties we require. In my intuition step, we simply pick one of those numbers. We are not making some unjustified assumption that may be suggested by the word “hypothesis." We call the number that we pick k. So OBVIOUSLY the following statement must be true
[math]k \text { is an integer} \ge 0 \text { and } \sum_{i=0}^k (2i + 1)^2 = \dfrac{(k + 1)(2k + 1)(2k + 3)}{3}.[/math]
Now in the induction step, we merely prove that that fact is also true for k + 1, which obviously must be an integer too.
[math] k + 1 \text { is an integer} \ge 0 \text { and}\\
\sum_{i=0}^{k+1} \{2(i + 1) + 1\}^2 =
\left (\sum_{i=0}^k (2i + 1)^2 \right ) + \{2(k + 1) + 1)^2 =\\
\dfrac{(k + 1)(2k + 1)(2k + 3)}{3} + (2k + 3)^2 = \\
\dfrac{(2k^2 + 3k + 1)(2k + 3)}{3} + 4k^2 + 12k + 9 =\\
\dfrac{4k^3 +12k^2 + 11k + 3}{3} + \dfrac{12k^2 + 36k + 27}{3} = \\
\dfrac{4k^3 + 24k^2 + 47k + 30}{3} =\dfrac{(k + 2)(4k^2 + 16k + 15) }{3} =\\
\dfrac{(k + 2)(2k + 3)(2k + 5)}{3} = \dfrac{\{k + 1) + 1\}\{2(k + 1) + 1\}\{2(k + 1) + 3\}}{3}.
[/math]
We are done.
