What is the coefficient of x^5 in the expansion of (x - 3)^7?

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Question: What is the coefficient of x^5 in the expansion of (x - 3)^7?

Options:

Correct Answer: -63

Solution:

The coefficient of x^5 is C(7,5) * (-3)^2 = 21 * 9 = -189.

What is the coefficient of x^5 in the expansion of (x - 3)^7?

What is the coefficient of x^5 in the expansion of (x - 3)^7?

Step 1: Identify the expression we need to expand, which is (x - 3)^7.
Step 2: Use the Binomial Theorem, which states that (a + b)^n = Σ [C(n, k) * a^(n-k) * b^k] for k = 0 to n.
Step 3: In our case, a = x, b = -3, and n = 7.
Step 4: We want the coefficient of x^5, which means we need to find the term where the power of x is 5.
Step 5: To find this term, we set n - k = 5, which means k = 7 - 5 = 2.
Step 6: Calculate the binomial coefficient C(7, 2), which is the number of ways to choose 2 from 7.
Step 7: C(7, 2) = 7! / (2! * (7-2)!) = 7! / (2! * 5!) = (7 * 6) / (2 * 1) = 21.
Step 8: Now, calculate (-3)^2, which is 9.
Step 9: Multiply the coefficient C(7, 2) by (-3)^2: 21 * 9 = 189.
Step 10: Since we are looking for the coefficient of x^5 in (x - 3)^7, we take the negative sign from (-3), so the final coefficient is -189.

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