What is the coefficient of x^5 in the expansion of (x + 1/2)^10?

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Q: What is the coefficient of x^5 in the expansion of (x + 1/2)^10?

Solution: The coefficient of x^5 is C(10,5) * (1/2)^5 = 252 * 1/32 = 7.875.

Steps: 10

Step 1: Identify the expression we are expanding, which is (x + 1/2)^10.
Step 2: Recognize that we want the coefficient of x^5 in this expansion.
Step 3: 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 4: In our case, a = x, b = 1/2, and n = 10.
Step 5: We need to find the term where x is raised to the power of 5, which means k = 10 - 5 = 5.
Step 6: Calculate the binomial coefficient C(10, 5), which is the number of ways to choose 5 items from 10.
Step 7: C(10, 5) = 10! / (5! * (10-5)!) = 252.
Step 8: Now, calculate (1/2)^5, which is 1/32.
Step 9: Multiply the coefficient C(10, 5) by (1/2)^5: 252 * (1/32).
Step 10: Perform the multiplication: 252 / 32 = 7.875.