What is the coefficient of x^5 in the expansion of (x + 3)^8?

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Question: What is the coefficient of x^5 in the expansion of (x + 3)^8?

Options:

Correct Answer: 1680

Solution:

Using the binomial theorem, the coefficient of x^5 in (x + 3)^8 is given by 8C5 * (3)^3 = 56 * 27 = 1512.

What is the coefficient of x^5 in the expansion of (x + 3)^8?

What is the coefficient of x^5 in the expansion of (x + 3)^8?

Step 1: Identify the expression we need to expand, which is (x + 3)^8.
Step 2: Recall the binomial theorem, which states that (a + b)^n = sum of (nCk * a^(n-k) * b^k) for k from 0 to n.
Step 3: In our case, a = x, b = 3, and n = 8.
Step 4: We want the coefficient of x^5, which means we need to find the term where x is raised to the power of 5.
Step 5: To find this term, we set n - k = 5, which means k = 8 - 5 = 3.
Step 6: Now, we need to calculate the binomial coefficient 8C3, which is the number of ways to choose 3 items from 8.
Step 7: Calculate 8C3 using the formula nCk = n! / (k! * (n-k)!), which gives us 8C3 = 8! / (3! * 5!) = 56.
Step 8: Next, we need to calculate (3)^k, where k = 3. So, (3)^3 = 27.
Step 9: Now, multiply the coefficient we found (56) by (3)^3 (which is 27): 56 * 27.
Step 10: Calculate 56 * 27 to get the final coefficient: 1512.

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