What is the value of the coefficient of x^5 in the expansion of (x + 1)^8?

What is the value of the coefficient of x^5 in the expansion of (x + 1)^8?

Step 1: Understand that we need to find the coefficient of x^5 in the expression (x + 1)^8.
Step 2: Recognize that (x + 1)^8 can be expanded using 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 = 1, and n = 8.
Step 4: We want the term where x is raised to the power of 5, which means we need to find the term where k = 3 (because 8 - 5 = 3).
Step 5: Use the binomial coefficient C(8, 3) to find the coefficient of the term with x^5.
Step 6: Calculate C(8, 3) using the formula C(n, k) = n! / (k! * (n-k)!). Here, C(8, 3) = 8! / (3! * 5!).
Step 7: Calculate 8! = 40320, 3! = 6, and 5! = 120.
Step 8: Now calculate C(8, 3) = 40320 / (6 * 120) = 40320 / 720 = 56.
Step 9: Therefore, the coefficient of x^5 in the expansion of (x + 1)^8 is 56.

No concepts available.