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Whats the coefficient of x^3 the expansion of (2x - 4)^7

User AVC
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The given form to expand is (2x-4)^7

and we need to find the coefficient of x^3.

The binomial formula is


(a+b)^n=\sum_{r\mathop{=}0}^nC_r^na^(n-r)b^r

where n is a positive integer and a, b are real numbers, and 0 < r ≤ n

Substituting the values a = 2x, b = -4, n = 7 and r = 4

Then we have,


(2x-4)^7=C_4^7(2x)^3(-4)^4

Now the coefficient of x^3 will be


\begin{gathered} C_4^7*8x^3*256 \\ =(7!)/(4!3!)*8*256* x^3 \\ =(7*6*5)/(3*2*1)*8*256* x^3 \\ =35*8*256* x^3 \\ =71680x^3 \end{gathered}

Hence, the coefficient will be 71680.

User SeGa
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