How to expand binomials? I'm working on a few proofs and am missing how this algebra works....
So, how does one expand $(k+1)^3\,$? Can I use FOIL? What does it expand to?
And how to expand $(k+1)^5\,$? 
Thanks!
 A: You can use "FOIL" twice.  You should get
$$
k^3+3k^2+3k+1.
$$
More generally,
$$
(a+b)^3 = a^3 + 3a^2b+3ab^2+ b^3.
$$
Please don't vacilate between lower-case $k$ and capital $K$ in mathematical notation.  Pick one and stick to it.  Mathematical notation is case sensitive.  Sometimes one uses lower-case $k$ and capital $K$ for two different things in the same problem, and you need to be clear about which is which.
But it would take a while to use FOIL to get
$$
(a+b)^9 = a^9+9a^8b+36a^7b^2+84a^6b^3+126a^5b^4+126a^4b^5+84a^3b^6+36a^2b^7+9ab^8+b^9.
$$
That's one reason to be aware of the binomial theorem, which explains the pattern.
A: Before you jump to the binomial theorem (still the best way to go, in general, for expressions of the form $(a+b)^n$), let's start at the beginning. You undoubtedly know that 
$$
(k+1)^3=(k+1)(k+1)(k+1)
$$
We'll start by expanding $(k+1)(k+1)$. You can use FOIL here so we have
$$
(k+1)(k+1)=k^2+2k+1
$$
We're two-thirds of the way to the answer. Now we have
$$
(k+1)^3=(k+1)(k^2+2k+1)
$$
and by the distributive property, namely that $(a+b)c=ac+bc$, we have
$$
\begin{align}
(k+1)^3=(k+1)(k^2+2k+1)&=(k)(k^2+2k+1)+(1)(k^2+2k+1)\\
               &=(k^3+2k^2+k)+(k^2+2k+1)\\
               &=k^3+3k^2+3k+1
\end{align}
$$
This will work for any positive integer exponent but, as Michael notes, you wouldn't want to do this for $(a+b)^n$.
A: Check out the entry Binomial Theorem in Wikipedia.
Putting $y = 1$, that will give you the tools you need to expand $(k+1)^3,\; (k+1)^5, \;$ and $\,(k + 1)^n\,$ for any non-negative integer $n$. 
FOIL works fine for $(k + 1)^2 = k^2 + 2k + 1$
One can go a step further by distributing $(k+1)$ over $(k^2 + 2k + 1)$ to get $$(k^3 + 3k^2 + 3k + 1) = (k + 1)^3.$$
But for large exponents, it's handy to know the pattern of coefficients that correspond to different powers of $k$ in the expansion of $(k+1)^n$: Pascal's triangle shows this handy relationship.
I'll include an animation and image of "Pascal's Triangle" which displays the coefficients of expansions of a binomial $(k + 1)$ (these coefficients are referred to as: binomial coefficients): up to and including fourth and fifth degree binomials, respectively:
$\quad\quad\quad\quad$ $\quad\quad\quad\quad$
$$\text{Each number in the triangle is the sum of the two directly above it.}$$

To see how this "plays out" in the expansion of $(x + 1)^n,\;0 \le n \le 6$:
$$(x + 1)^0 = \color{blue}{\bf{1}}$$
$$(x + 1)^1 = \color{blue}{\bf{1}}\cdot x +\color{blue}{\bf{1}}$$
$$(x + 1)^2 = \color{blue}{\bf{1}}\cdot x^2 + \color{blue}{\bf{2}}x + \color{blue}{\bf{1}}$$
$$(x+1)^3 = \color{blue}{\bf{1}}\cdot x^3 + \color{blue}{\bf{3}}x^2 + \color{blue}{\bf{3}}x + \color{blue}{\bf{1}}$$
$$(x+1)^4 = \color{blue}{\bf{1}}\cdot x^4 + \color{blue}{\bf{4}} x^3+ \color{blue}{\bf{6}}x^2 + \color{blue}{\bf{4}}x +\color{blue}{\bf{1}}$$
$$(x+1)^5 = \color{blue}{\bf{1}}\cdot x^5 + \color{blue}{\bf{5}}x^4 + \color{blue}{\bf{10}} x^3 + \color{blue}{\bf{10}} x^2 + \color{blue}{\bf{5}}x + \color{blue}{\bf{1}}$$
$$(x + 1)^6 = \color{blue}{\bf{1}}\cdot x^6 + \color{blue}{\bf{6}}x^5 +\color{blue}{\bf{15}}x^4 + \color{blue}{\bf{20}}x^3 +\color{blue}{\bf{15}}x^2 + \color{blue}{\bf{6}}x + \color{blue}{\bf{1}}$$
$${\bf{\vdots}}$$
A: What you're looking for is the binomial theorem, where y = 1. 
