How do you find the fourth term of ${({(2 x - z)}^{2})}^{6}$ $((2x-z)^2 )^6$ ?

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How do you find the fourth term of ${({(2 x - z)}^{2})}^{6}$ $((2x-z)^2 )^6$ ?

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Answer 1 · 2015-05-27T23:44:17

${({(2 x - z)}^{2})}^{6} = {(2 x - z)}^{12}$ $((2x-z)^2)^6 = (2x-z)^12$

The fourth term of this expansion is
$12 C 4 \cdot {(2 x)}^{4} \cdot {(- z)}^{12 - 4}$ $color(red)(12C4) * (2x)^4 * (-z)^(12-4)$

$= \frac{12!}{(4!) (8!)} \cdot 16 x^{4} \cdot z^{8}$ $= (12!)/((4!)(8!)) * 16x^4 * z^8$

$= (495) \cdot 16 x^{4} z^{8}$ $= (495) *16x^4z^8$

$= 7920 x^{4} z^{8}$ $=7920x^4z^8$

$Note$ $color(red)("Note")$
I'm not certain of the current notation standard for this expression: "12 Choose 4" and have used the form I learned more than 50 years ago.

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How do you find the fourth term of ${({(2 x - z)}^{2})}^{6}$ $((2x-z)^2 )^6$ ?

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How do you find the fourth term of ((2x-z)^2 )^6((2x−z)2)6((2x-z)^2 )^6?

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How do you find the fourth term of ${({(2 x - z)}^{2})}^{6}$ $((2x-z)^2 )^6$ ?