What is the coefficient of x^2 in the expansion of (2x - 5)^5? (2019)

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Question: What is the coefficient of x^2 in the expansion of (2x - 5)^5? (2019)

Options:

Correct Answer: -600

Exam Year: 2019

Solution:

The coefficient of x^2 in (2x - 5)^5 is given by 5C2 * (2)^2 * (-5)^3 = 10 * 4 * (-125) = -5000.

What is the coefficient of x^2 in the expansion of (2x - 5)^5? (2019)

What is the coefficient of x^2 in the expansion of (2x - 5)^5? (2019)

Step 1: Identify the expression we need to expand, which is (2x - 5)^5.
Step 2: Recognize that we want the coefficient of x^2 in this expansion.
Step 3: Use the Binomial Theorem, which states that (a + b)^n = sum of (nCk * a^(n-k) * b^k) for k from 0 to n.
Step 4: In our case, a = 2x, b = -5, and n = 5.
Step 5: We need to find the term where the power of x is 2. This happens when we choose (2x) twice and (-5) three times.
Step 6: Calculate the number of ways to choose 2 from 5, which is 5C2. This is equal to 5! / (2!(5-2)!) = 10.
Step 7: Calculate (2)^2, which is 4.
Step 8: Calculate (-5)^3, which is -125.
Step 9: Multiply these values together: 5C2 * (2)^2 * (-5)^3 = 10 * 4 * (-125).
Step 10: Perform the multiplication: 10 * 4 = 40, and then 40 * (-125) = -5000.
Step 11: Conclude that the coefficient of x^2 in the expansion of (2x - 5)^5 is -5000.

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