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Chapter 25: Problem 730

Find the expansion of \((a-2 x)^{7}\).

Short Answer

Expert verified
The expansion of \((a-2x)^7\) is \(a^7 - 14a^6x + 84a^5x^2 - 280a^4x^3 + 560a^3x^4 - 672a^2x^5 + 448ax^6 - 128x^7\).

Step by step solution

01

Identify a, b, and n

In this case, our values of a, b, and n are as follows: a = \(a\) b = \(-2x\) n = 7
02

Apply the Binomial Theorem

Let's apply the Binomial Theorem to expand \((a - 2x)^7\). According to the theorem, the expansion is given by: \[(a - 2x)^7 = \sum_{k=0}^{7} \binom{7}{k} a^{7-k}(-2x)^{k}\]
03

Expand Each Term

Now we will expand each term in the sum: k = 0: \[\binom{7}{0}a^{7-0}(-2x)^{0} = 1\cdot a^7\cdot 1 = a^7\] k = 1: \[\binom{7}{1}a^{7-1}(-2x)^{1} = 7\cdot a^6\cdot (-2x) = -14a^6x\] k = 2: \[\binom{7}{2}a^{7-2}(-2x)^{2} = 21\cdot a^5\cdot 4x^2 = 84a^5x^2\] k = 3: \[\binom{7}{3}a^{7-3}(-2x)^{3} = 35\cdot a^4\cdot (-8x^3) = -280a^4x^3\] k = 4: \[\binom{7}{4}a^{7-4}(-2x)^{4} = 35\cdot a^3\cdot 16x^4 = 560a^3x^4\] k = 5: \[\binom{7}{5}a^{7-5}(-2x)^{5} = 21\cdot a^2\cdot (-32x^5) = -672a^2x^5\] k = 6: \[\binom{7}{6}a^{7-6}(-2x)^{6} = 7\cdot a\cdot 64x^6 = 448ax^6\] k = 7: \[\binom{7}{7}a^{7-7}(-2x)^{7} = 1\cdot 1\cdot (-128x^7) = -128x^7\]
04

Sum the Terms

Now we sum the terms to obtain the expanded form of \((a - 2x)^7\): \((a - 2x)^7 = a^7 - 14a^6x + 84a^5x^2 - 280a^4x^3 + 560a^3x^4 - 672a^2x^5 + 448ax^6 - 128x^7\) The expansion of \((a-2x)^7\) is \(a^7 - 14a^6x + 84a^5x^2 - 280a^4x^3 + 560a^3x^4 - 672a^2x^5 + 448ax^6 - 128x^7\).

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