What is the coefficient of `x^(3) y^(4) `in `( 2x + 3y^(2) ) ^(5)` ?

Options

A. 240
B. 360
C. 720
D. 1080

Answer

To find the coefficient of \( x^3 y^4 \) in the expression \( (2x + 3y^2)^5 \), we can use the Binomial Theorem. The Binomial Theorem states that:

\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

In our case, \( a = 2x \), \( b = 3y^2 \), and \( n = 5 \).

### Step 1: Identify the terms
We want to find the term in the expansion that contains \( x^3 \) and \( y^4 \).

### Step 2: Set up the powers
From the term \( (2x)^{n-k} \) and \( (3y^2)^k \), we need to determine \( n-k \) and \( k \) such that:
- The power of \( x \) is 3: \( n-k = 3 \)
- The power of \( y \) is 4: \( k \) must satisfy \( 2k = 4 \) (since \( y^2 \) is raised to the power \( k \))

### Step 3: Solve for \( k \)
From \( 2k = 4 \), we find:
\[ k = 2 \]

### Step 4: Solve for \( n-k \)
Now substituting \( k = 2 \) into \( n-k = 3 \):
\[ n - 2 = 3 \implies n = 5 \]
This is consistent since \( n = 5 \).

### Step 5: Calculate the binomial coefficient
Next, we calculate the binomial coefficient \( \binom{5}{2} \):
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

### Step 6: Calculate the term
Now we substitute \( k = 2 \) into the expression:
\[ (2x)^{5-2} = (2x)^3 = 8x^3 \]
\[ (3y^2)^2 = 9y^4 \]

### Step 7: Combine everything
Now we can find the coefficient of the term:
\[ \text{Coefficient} = \binom{5}{2} \times 8 \times 9 = 10 \times 8 \times 9 \]

Calculating this gives:
\[ 10 \times 8 = 80 \]
\[ 80 \times 9 = 720 \]

### Final Answer
Thus, the coefficient of \( x^3 y^4 \) in \( (2x + 3y^2)^5 \) is **720**.

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