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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:

enter image description here

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    $\begingroup$ What is the original question? If this comes up as part of a proof, there might be other material that matters. $\endgroup$ – Clayton Mar 8 '13 at 2:34
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    $\begingroup$ I've edited your question to use $\LaTeX$. Please make sure it still represents your original intent. For help with formatting for future questions, please see this (really neat) meta question. $\endgroup$ – apnorton Mar 8 '13 at 2:38
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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 $(2)$

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    $\begingroup$ sm 1: When you have the chance, perhaps when you're at a computer, if after checking and reworking your algebra, if you can't seem to find the error in your proof (which I imagine is happening in the inductive step), feel free to post a separate question, providing a bit more in the way of details: what you're trying to prove, what you've got, etc. and we can help out if needed. $\endgroup$ – Namaste Mar 8 '13 at 2:38
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What you need to show instead is $$\frac{((n+1)+1)(2(n+1)+1)(2(n+1)+3)}{3}-\frac{(n+1)(2n+1)(2n+3)}{3}=(2(n+1)+1)^2$$

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