What is the coefficient of x^5 in the expansion of (x + 2)^8?
Practice Questions
1 question
Q1
What is the coefficient of x^5 in the expansion of (x + 2)^8?
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The coefficient of x^5 is C(8,5) * (2)^3 = 56 * 8 = 448.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the coefficient of x^5 in the expansion of (x + 2)^8?
Solution: The coefficient of x^5 is C(8,5) * (2)^3 = 56 * 8 = 448.
Steps: 8
Step 1: Identify the expression to expand, which is (x + 2)^8.
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 = 2, and n = 8.
Step 4: We want the term where x has the power of 5, which means we need to find the term where k = 3 (since 8 - 5 = 3).
Step 5: Calculate the binomial coefficient C(8, 3), which is the number of ways to choose 3 items from 8. This is calculated as 8! / (3! * (8-3)!) = 56.
Step 6: Calculate 2^3, which is 2 * 2 * 2 = 8.
Step 7: Multiply the coefficient from Step 5 by the result from Step 6: 56 * 8 = 448.
Step 8: The coefficient of x^5 in the expansion of (x + 2)^8 is 448.
