Some things to note:
The sum of all the coefficients is \(2^{n}\). To prove this, evaluate \(\left(a+b\right)^{n}\) with \(a=1,b=1\)
The sum of the odd and even binomial coefficients is the same. To prove this, evaluate \(\left(a+b\right)^{n}\) with \(a=1,b=-1\)
\((1+x)^{n}\)
If \(|x|<1\), then:
\begin{equation*}
(1+x)^{n}=1+n x+\frac{n(n-1)}{2 !}+\frac{n(n-1)(n-2)}{3 !} x^{3}+\ldots
\end{equation*}
One interesting use for this is to approximate certain numbers as an infinte series:
Consider:
\begin{equation*}
1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot6}+\frac{1\cdot3 \cdot5}{3\cdot6\cdot9}+\ldots
\end{equation*}
Equate it term by term with the series expansion above to identify \(x\) and \(n\).
Important: Always cross check that \(|x|<1\) when you get your answer!
Another example:
Show that if:
\begin{equation*}
y=\frac{1}{2}\left(\frac{4}{9}\right)+\frac{1\cdot3}{2^{2} \cdot 2 !}\left(\frac{4}{9}\right)^{2}+\frac{1\cdot3 \cdot5}{2^{3} \cdot 3 !}\left(\frac{4}{9}\right)^{3}+\ldots
\end{equation*}
Then \(5 y^{2}+10 y-4=0\)
The trick here is to consider \(1+y\) and then do a term by term matching.