In the expansion of (x + 2)^7, what is the term containing x^5?

In the expansion of (x + 2)^7, what is the term containing x^5?

Step 1: Identify the expression we are expanding, which is (x + 2)^7.
Step 2: Use the Binomial Theorem, which states that (a + b)^n = Σ (nCk * a^(n-k) * b^k) for k = 0 to n.
Step 3: In our case, a = x, b = 2, and n = 7.
Step 4: We want the term that contains x^5. This means we need to find k such that n - k = 5.
Step 5: Solve for k: 7 - k = 5, which gives k = 2.
Step 6: Now, we can find the term using k = 2: The term is given by 7C2 * (x)^(7-2) * (2)^2.
Step 7: Calculate 7C2, which is 7! / (2! * (7-2)!) = 21.
Step 8: Calculate (2)^2, which is 4.
Step 9: Combine these results: The term is 21 * 4 * x^5.
Step 10: Multiply 21 and 4 to get 84, so the term is 84x^5.

No concepts available.