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

Options

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

Explanation

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

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

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

### Step 1: Identify the general term
The general term in the expansion of \( (2x + 3y^2)^5 \) is given by:

\[ T_r = \binom{5}{r} (2x)^{5-r} (3y^2)^r \]

### Step 2: Simplify the general term
We can simplify this term:

\[ T_r = \binom{5}{r} (2^{5-r} x^{5-r}) (3^r (y^2)^r) = \binom{5}{r} 2^{5-r} 3^r x^{5-r} y^{2r} \]

### Step 3: Determine the values of \( r \)
We need to find the values of \( r \) such that the term includes \( x^3 \) and \( y^4 \). This gives us two equations:

1. From \( x^{5-r} = x^3 \), we get:
\[ 5 - r = 3 \implies r = 2 \]

2. From \( y^{2r} = y^4 \), we get:
\[ 2r = 4 \implies r = 2 \]

Thus, both conditions are satisfied when \( r = 2 \).

### Step 4: Substitute \( r \) into the general term
Now we substitute \( r = 2 \) into the general term:

\[ T_2 = \binom{5}{2} 2^{5-2} 3^2 x^{5-2} y^{2 \cdot 2} \]

### Step 5: Calculate the coefficient
Calculating each part:

1. \( \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \)
2. \( 2^{5-2} = 2^3 = 8 \)
3. \( 3^2 = 9 \)

Putting it all together:

\[ T_2 = 10 \cdot 8 \cdot 9 \cdot x^3 \cdot y^4 \]

Calculating the coefficient:

\[ 10 \cdot 8 \cdot 9 = 720 \]

### Final Answer
The coefficient of \( x^3 y^4 \) in \( (2x + 3y^2)^5 \) is \( \boxed{720} \).

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