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

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

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Answer 1 · 2016-02-27T09:16:08

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

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

Explanation:

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

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

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

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

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Explanation:

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