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How do I expand the following? $(2k + 1)^2$

I'm an individual with extremely limited background knowledge of mathematics but they're making us expand this for discrete mathematics when creating 'proofs'.

How can I expand a number format like this and what should I learn to know how to do this in future? (Apologies for being silly)

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The straight-forward way: $$ (2k + 1)^2 = (2k+1)(2k+1)\\ = 2k(2k+1) + 1(2k+1)\\ = 4k^2 + 2k + 2k + 1\\ = 4k^2 + 4k + 1 $$


The shortcut: If you're already familiar with the formula $(a+b)^2 = a^2 + 2ab + b^2$ (which you can deduce on your own like we did above), we can do it in fewer steps by letting $2k$ fill in for $a$ and $1$ fill in for $b$: $$ (2k+1)^2 = (2k)^2 + 2\cdot 2k\cdot 1 + 1^2\\ = 4k^2 + 4k + 1 $$

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You can expand the expression of interest by a direct computation $$(2k+1)^2 = (2k+1)(2k+1) = 4k^2 + 2k +2k +1 = 4k^2 + 4k + 1$$ In general, one finds that $$ (a+b)^2 = a^2 + b^2 + 2ab$$ Expression such as $$ (a+b)^n $$ can be be computed using the Binomial Theorem https://en.wikipedia.org/wiki/Binomial_theorem, if you wish to spare the effort of a direct calculation.

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The following image ... dumbly dumbly doo ... enter image description here

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There are two possibilities. There is a formula and there are the basic rules of expansion using distributivity of the multiplication over addition.

1) Apply the formula $(a+b)^2=a^2+2ab+b^2$ with $a=2k$ and $b=1$. You can prove this formula easily by writing $(a+b)^2=(a+b)(a+b)$ and expanding using the other formula $$(a+b)\times(c+d)=a\times c+a\times d+b\times c+b\times d\,.$$

2) The other method is to do like the proof of the formula I mentioned in 1). You simply write $$(2k+1)²=(2k+1)\times(2k+1)$$ and then you start expanding using distributivity $$(2k+1)\times (2k+1)=2k\times 2k+2k\times 1+\dots$$

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The binomial theorem says

$$ (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k $$

where $\binom{m}{n} = \frac{m!}{n!(m-n)!}$.

For your example, $(2k+1)^2$, where $a=2k$, $b=1$, and $n=2$:

\begin{align*} (2k+1)^2 &= \binom{2}{0} (2k)^{2-0}1^0 + \binom{2}{1} (2k)^{2-1}1^1 + \binom{2}{2} (2k)^{2-2}1^2 \\ &= 4k^2 + 4k + 1 \end{align*}

This, however, is kind of overkill for a squared binomial as we have the general form, $(a+b)^2 = a^2 + 2ab + b^2$.

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