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Answer :
To expand [tex]\((2x - 1)^6\)[/tex], we'll use the Binomial Theorem. The Binomial Theorem states that:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
In this case, [tex]\(a = 2x\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(n = 6\)[/tex]. We will expand each term using the formula:
[tex]\[
\sum_{k=0}^{6} \binom{6}{k} (2x)^{6-k} (-1)^k
\][/tex]
Let's expand each of these:
1. For [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{6}{0} (2x)^6 (-1)^0 = 1 \cdot (64x^6) = 64x^6
\][/tex]
2. For [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{6}{1} (2x)^5 (-1)^1 = 6 \cdot 32x^5 \cdot (-1) = -192x^5
\][/tex]
3. For [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{6}{2} (2x)^4 (-1)^2 = 15 \cdot 16x^4 = 240x^4
\][/tex]
4. For [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{6}{3} (2x)^3 (-1)^3 = 20 \cdot 8x^3 \cdot (-1) = -160x^3
\][/tex]
5. For [tex]\(k = 4\)[/tex]:
[tex]\[
\binom{6}{4} (2x)^2 (-1)^4 = 15 \cdot 4x^2 = 60x^2
\][/tex]
6. For [tex]\(k = 5\)[/tex]:
[tex]\[
\binom{6}{5} (2x)^1 (-1)^5 = 6 \cdot 2x \cdot (-1) = -12x
\][/tex]
7. For [tex]\(k = 6\)[/tex]:
[tex]\[
\binom{6}{6} (2x)^0 (-1)^6 = 1 \cdot 1 = 1
\][/tex]
Adding all these terms together gives us the expanded expression:
[tex]\[
64x^6 - 192x^5 + 240x^4 - 160x^3 + 60x^2 - 12x + 1
\][/tex]
Therefore, the correct choice is:
D. [tex]\(64 x^6 - 192 x^5 + 240 x^4 - 160 x^3 + 60 x^2 - 12 x + 1\)[/tex]
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
In this case, [tex]\(a = 2x\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(n = 6\)[/tex]. We will expand each term using the formula:
[tex]\[
\sum_{k=0}^{6} \binom{6}{k} (2x)^{6-k} (-1)^k
\][/tex]
Let's expand each of these:
1. For [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{6}{0} (2x)^6 (-1)^0 = 1 \cdot (64x^6) = 64x^6
\][/tex]
2. For [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{6}{1} (2x)^5 (-1)^1 = 6 \cdot 32x^5 \cdot (-1) = -192x^5
\][/tex]
3. For [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{6}{2} (2x)^4 (-1)^2 = 15 \cdot 16x^4 = 240x^4
\][/tex]
4. For [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{6}{3} (2x)^3 (-1)^3 = 20 \cdot 8x^3 \cdot (-1) = -160x^3
\][/tex]
5. For [tex]\(k = 4\)[/tex]:
[tex]\[
\binom{6}{4} (2x)^2 (-1)^4 = 15 \cdot 4x^2 = 60x^2
\][/tex]
6. For [tex]\(k = 5\)[/tex]:
[tex]\[
\binom{6}{5} (2x)^1 (-1)^5 = 6 \cdot 2x \cdot (-1) = -12x
\][/tex]
7. For [tex]\(k = 6\)[/tex]:
[tex]\[
\binom{6}{6} (2x)^0 (-1)^6 = 1 \cdot 1 = 1
\][/tex]
Adding all these terms together gives us the expanded expression:
[tex]\[
64x^6 - 192x^5 + 240x^4 - 160x^3 + 60x^2 - 12x + 1
\][/tex]
Therefore, the correct choice is:
D. [tex]\(64 x^6 - 192 x^5 + 240 x^4 - 160 x^3 + 60 x^2 - 12 x + 1\)[/tex]
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