In the expansion of (1 + x)^10, what is the coefficient of x^5?

1 question

1 item

Q: In the expansion of (1 + x)^10, what is the coefficient of x^5?

Solution: The coefficient of x^5 is C(10,5) = 10! / (5!5!) = 252.

Steps: 10

Step 1: Understand that (1 + x)^10 is a binomial expression that can be expanded using the Binomial Theorem.
Step 2: The Binomial Theorem states that (a + b)^n = Σ (n choose k) * a^(n-k) * b^k, where k goes from 0 to n.
Step 3: In our case, a = 1, b = x, and n = 10.
Step 4: We want to find the coefficient of x^5 in the expansion, which corresponds to k = 5.
Step 5: The coefficient of x^5 is given by the binomial coefficient C(10, 5).
Step 6: Calculate C(10, 5) using the formula C(n, k) = n! / (k!(n-k)!). Here, n = 10 and k = 5.
Step 7: Substitute the values into the formula: C(10, 5) = 10! / (5! * 5!).
Step 8: Calculate 10! = 3628800, 5! = 120, so C(10, 5) = 3628800 / (120 * 120).
Step 9: Simplify the calculation: 120 * 120 = 14400, then 3628800 / 14400 = 252.
Step 10: Therefore, the coefficient of x^5 in the expansion of (1 + x)^10 is 252.