What is the coefficient of x^0 in the expansion of (x + 5)^5?

What is the coefficient of x^0 in the expansion of (x + 5)^5?

Correct Answer: 3125

Step 1: Understand that x^0 means we are looking for the constant term in the expansion of (x + 5)^5.
Step 2: Recall the binomial theorem, which states that (a + b)^n can be expanded using the formula: C(n, k) * a^(n-k) * b^k, where C(n, k) is the binomial coefficient.
Step 3: In our case, a = x, b = 5, and n = 5.
Step 4: To find the coefficient of x^0, we need to set k = 5 (because we want all of the x's to disappear).
Step 5: Calculate the binomial coefficient C(5, 5), which is equal to 1 (there's only one way to choose all 5 items).
Step 6: Now calculate 5^5, which is 5 multiplied by itself 5 times: 5 * 5 * 5 * 5 * 5 = 3125.
Step 7: Multiply the binomial coefficient by 5^5: 1 * 3125 = 3125.
Step 8: Conclude that the coefficient of x^0 in the expansion of (x + 5)^5 is 3125.

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