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#?

1 Answer
Feb 27, 2016

Answer:

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}#