How would this:
$$\frac{((n+1)+1)(2(n+1) + 1)(2(n+1) + 3)}{3}$$
Factor to this:
$$(2(n+1)+1)^2$$
This is a part of an induction proof, which I would post an image if my reputation was higher... Writing this from a phone but will add more detail later if needed thanks again.
Thank you, please include all the steps-
EDIT Thanks amWhy, must be something else happening in the proof...
EDIT Thanks anorton the LaTex helps, so below is the link to the proof, I'm still trying to understand the initial inductive steps in this method, let me know if more is needed:
Your first equation, unless there's a typo, will not factor into $[2(n+1)+1]^2$:
Your first equation is a cubic polynomial (degree 3), and the second is a quadratic, a degree two polynomial, so they cannot be equivalent.
*Edit: (given image of problem statement):
Given your hypothesis of assuming the truth that $$\sum_{j=0}^n (2j+1)^2 = \frac{(n+1)(2n + 1)(2n + 3)}{3} \;\;\;\text{holds for $n$,}$$
then what you need to show, having used your inductive hypothesis correctly, is that
$$\frac{(n+1)(2n + 1)(2n + 3)}{3} + [2(n+1) = 1]^2 = \frac{((n+1)+1)(2(n+1) + 1)(2(n+1) + 3)}{3}\quad\quad\quad\tag{1}$$
Or, alternatively, and more straightforward, we can prove $(1)$ by showing that:
$$\frac{(n+1)(2n + 1)(2n + 3)}{3} -\frac{((n+1)+1)(2(n+1) + 1)(2(n+1) + 3)}{3} = [2(n+1) = 1]^2\quad\quad\quad\tag{2}$$
And proving equation $(2)$ should proceed smoothly, if you're careful with your algebra! The terms involving $n^3$ will cancel out, etc, and you'll end with the quadratic on the right-hand side of equation $(2)$