In the expansion of (2 + 3x)^5, what is the coefficient of x^2?

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Question: In the expansion of (2 + 3x)^5, what is the coefficient of x^2?

Options:

Correct Answer: 180

Solution:

The coefficient of x^2 is given by 5C2 * (3^2) * (2^3) = 10 * 9 * 8 = 720.

In the expansion of (2 + 3x)^5, what is the coefficient of x^2?

In the expansion of (2 + 3x)^5, what is the coefficient of x^2?

Step 1: Identify the expression to expand, which is (2 + 3x)^5.
Step 2: Recognize that we need to find the coefficient of x^2 in the expansion.
Step 3: Use the binomial theorem, which states that (a + b)^n = Σ (nCk * a^(n-k) * b^k) for k = 0 to n.
Step 4: In our case, a = 2, b = 3x, and n = 5.
Step 5: We want the term where x has the power of 2, which means we need k = 2.
Step 6: Calculate nCk, which is 5C2. This is the number of ways to choose 2 from 5, calculated as 5! / (2!(5-2)!).
Step 7: Calculate 5C2 = 5! / (2! * 3!) = (5 * 4) / (2 * 1) = 10.
Step 8: Calculate (3x)^2, which is (3^2) * (x^2) = 9 * x^2.
Step 9: Calculate (2)^(5-2) = (2^3) = 8.
Step 10: Multiply the results: Coefficient = 5C2 * (3^2) * (2^3) = 10 * 9 * 8.
Step 11: Calculate 10 * 9 = 90, then 90 * 8 = 720.
Step 12: Conclude that the coefficient of x^2 in the expansion of (2 + 3x)^5 is 720.

Binomial Expansion – The question tests the understanding of the binomial theorem, specifically how to find coefficients in the expansion of a binomial expression.
Combinatorics – The use of binomial coefficients (nCr) to determine the number of ways to choose terms from the expansion.
Exponents and Powers – Understanding how to apply powers to constants and variables in the context of the binomial expansion.