What is the value of the 3rd term in the expansion of (3x - 4)^5? (2020)

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Question: What is the value of the 3rd term in the expansion of (3x - 4)^5? (2020)

Options:

Correct Answer: -240

Exam Year: 2020

Solution:

The 3rd term is C(5,2) * (3x)^3 * (-4)^2 = 10 * 27x^3 * 16 = -240.

What is the value of the 3rd term in the expansion of (3x - 4)^5? (2020)

What is the value of the 3rd term in the expansion of (3x - 4)^5? (2020)

Step 1: Identify the expression to expand, which is (3x - 4)^5.
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 = 3x, b = -4, and n = 5.
Step 4: We want to find the 3rd term in the expansion. The 3rd term corresponds to k = 2 (since we start counting from k = 0).
Step 5: Calculate C(5, 2), which is the number of combinations of 5 items taken 2 at a time. C(5, 2) = 5! / (2!(5-2)!) = 10.
Step 6: Calculate (3x)^(5-2) = (3x)^3 = 27x^3.
Step 7: Calculate (-4)^2 = 16.
Step 8: Combine these results to find the 3rd term: 10 * 27x^3 * 16.
Step 9: Multiply the numbers: 10 * 27 = 270, and then 270 * 16 = 4320.
Step 10: The final result for the 3rd term is 4320x^3.

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