# How do you use the binomial formula to find the coefficient of the z^2 x^10 term in the expansion of (2x+z)^12?

Feb 27, 2016

The coefficient of ${z}^{2} {x}^{10}$ is :

2^{10}C(12,2) = 2^{10}\frac{12!}{10!2!}=66\times2^{10}

#### Explanation:

Binomial Theorem: ${\left(a + b\right)}^{n} = \setminus {\sum}_{k = 0}^{n} C \left(n , k\right) {a}^{n - k} {b}^{k}$ where the binomial coefficient C(n,k)=\frac{n!}{k!(n-k)!}

${\left(2 x + z\right)}^{12} = \setminus {\sum}_{k = 0}^{12} C \left(12 , k\right) {\left(2 x\right)}^{12 - k} {z}^{k}$

Consider the $k = 2$ term : $C \left(12 , 2\right) {\left(2 x\right)}^{10} {z}^{2} = C \left(12 , 2\right) {2}^{10} {x}^{10} {z}^{2}$
So the coefficient of ${z}^{2} {x}^{10}$ is 2^{10}C(12,2) = 2^{10}\frac{12!}{10!2!}=66\times2^{10}