Find the coefficient of x^6 in the expansion of (2x+1)^12

Thanks Heureka,

you don't really need to do all that.

$$\\(2x+1)^{12}=\displaystyle\sum_{r=0}^{12}\;^{12}C_r(2x)^r(1)^{12-r}=\displaystyle\sum_{r=0}^{12}\;^{12}C_r(2x)^r\\\\
$The only term with x^6\; is $\\\\
^{12}C_6(2x)^6=^{12}C_6*2^6x^6\\\\
$So the coefficient of $x^6\;\;is \;\;\;^{12}C_6*2^6=^{12}C_6*64$$

$${\left({\frac{{\mathtt{12}}{!}}{{\mathtt{6}}{!}{\mathtt{\,\times\,}}({\mathtt{12}}{\mathtt{\,-\,}}{\mathtt{6}}){!}}}\right)}{\mathtt{\,\times\,}}{\mathtt{64}} = {\mathtt{59\,136}}$$

Find the coefficient of x^6 in the expansion of (2x+1)^12 

$$\\\small{\text{$(2x+1)^{12} $}}\\\\
\small{\text{
$
= \binom{12}{0}\cdot(2x)^{12}\cdot 1^0
+ \binom{12}{1}\cdot(2x)^{11}\cdot 1^1
+ \binom{12}{2}\cdot(2x)^{10}\cdot 1^2
+ \binom{12}{3}\cdot(2x)^{9}\cdot 1^3
+ \binom{12}{4}\cdot(2x)^{8}\cdot 1^4+
$}}\\\\
\small{\text{
$
+ \binom{12}{5}\cdot(2x)^{7}\cdot 1^5
+ \textcolor[rgb]{1,0,0}{
\binom{12}{6}\cdot(2x)^{6}\cdot 1^6
}
+ \binom{12}{7}\cdot(2x)^{5}\cdot 1^7
+ \binom{12}{8}\cdot(2x)^{4}\cdot 1^8+
$}}\\\\
\small{\text{
$
+ \binom{12}{9}\cdot(2x)^{3}\cdot 1^9
+ \binom{12}{10}\cdot(2x)^{2}\cdot 1^{10}
+ \binom{12}{11}\cdot(2x)^{1}\cdot 1^{11}
+ \binom{12}{12}\cdot(2x)^{0}\cdot 1^{12}
$}}$$

Coefficient of $$\small{\text{$x^6$}}$$ :

$$\small{\text{
$
\begin{array}{rl}
&\textcolor[rgb]{1,0,0}{\binom{12}{6}\cdot(2x)^{6}\cdot 1^6} \\\\
=&\binom{12}{6}\cdot(2x)^{6}\\\\
=&\binom{12}{6}\cdot2^6\cdot x^6\\\\
=&\binom{12}{6}\cdot 64 \cdot x^6\\\\
=&924\cdot 64 \cdot x^6\\\\
=&59136\cdot x^6\\\\
\end{array}
$
}}$$

The coefficient of $$\small{\text{$x^6$}}$$ is 59136