In the expansion of (3x - 4)^7, what is the coefficient of x^5? (1920)

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Q: In the expansion of (3x - 4)^7, what is the coefficient of x^5? (1920)

Solution: Using the binomial theorem, the coefficient of x^5 in (3x - 4)^7 is given by 7C5 * (3)^5 * (-4)^2 = 21 * 243 * 16 = 21 * 3888 = 81588.

Steps: 13

Step 1: Identify the expression we are working with, which is (3x - 4)^7.
Step 2: Recognize that we need to find the coefficient of x^5 in this expansion.
Step 3: Use the binomial theorem, which states that (a + b)^n = sum of (nCk * a^(n-k) * b^k) for k from 0 to n.
Step 4: In our case, a = 3x, b = -4, and n = 7.
Step 5: We want the term where x has the power of 5, which means we need to find the term where (3x)^(5) and (-4)^(2) are used.
Step 6: Calculate k, which is n - 5 = 7 - 5 = 2. So we will use k = 2.
Step 7: Calculate the binomial coefficient 7C2, which is 7! / (2!(7-2)!) = 21.
Step 8: Calculate (3)^5, which is 243.
Step 9: Calculate (-4)^2, which is 16.
Step 10: Multiply the results: 21 * 243 * 16.
Step 11: First, calculate 21 * 243 = 5103.
Step 12: Then, calculate 5103 * 16 = 81648.
Step 13: The coefficient of x^5 in the expansion of (3x - 4)^7 is 81648.