Question

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`?

Answer 1

The coefficient of `z^2x^10` is :

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

Explanation:

Binomial Theorem: `(a+b)^n=\sum_{k=0}^{n}C(n,k)a^{n-k}b^k` where the binomial coefficient `C(n,k)=\frac{n!}{k!(n-k)!}`

`(2x+z)^{12} = \sum_{k=0}^{12} C(12,k)(2x)^{12-k}z^k`

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