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