What is the coefficient of x^5 in the expansion of (x + 4)^7? (2020)

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Question: What is the coefficient of x^5 in the expansion of (x + 4)^7? (2020)

Options:

Correct Answer: 1680

Exam Year: 2020

Solution:

The coefficient of x^5 is C(7,5) * (4)^2 = 21 * 16 = 336.

What is the coefficient of x^5 in the expansion of (x + 4)^7? (2020)

What is the coefficient of x^5 in the expansion of (x + 4)^7? (2020)

Step 1: Identify the expression to expand, which is (x + 4)^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 = 4, and n = 7.
Step 4: We want the term where x has the power of 5, which means we need k = 2 (because 7 - 5 = 2).
Step 5: Calculate the binomial coefficient C(7, 5), which is the same as C(7, 2). This is calculated as 7! / (5! * 2!) = 21.
Step 6: Calculate 4^2, which is 16.
Step 7: Multiply the coefficient from Step 5 by the result from Step 6: 21 * 16 = 336.
Step 8: The coefficient of x^5 in the expansion of (x + 4)^7 is 336.

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