# How do you find the coefficient of a of the term ax^5 in the expansion of the binomial (x+3)^12?

Jun 18, 2017

The coefficient of ${x}^{5}$ is $1732104$

#### Explanation:

the term of ${x}^{r}$ in the expansion of ${\left(x + a\right)}^{n}$ is

${C}_{r}^{n} {x}^{r} {a}^{n - r}$

Hence, the term containing ${x}^{5}$ in the expansion of ${\left(x + 3\right)}^{12}$ is

${C}_{5}^{12} {x}^{5} {3}^{7} = \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} {x}^{5} \times 2187$

= $792 \times 2187$

= $1732104$