Find the coefficient of `x^(6).y^(3)` in the expansion of `(2x+y)^(9)`

Explanation

To find the coefficient of \( x^6 y^3 \) in the expansion of \( (2x + y)^9 \), we can use the Binomial Theorem. The Binomial Theorem states that:

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

In our case, \( p = 2x \), \( q = y \), and \( n = 9 \).

### Step 1: Identify the General Term
The general term \( T_{r+1} \) in the expansion of \( (2x + y)^9 \) is given by:

\[ T_{r+1} = \binom{9}{r} (2x)^{9-r} y^r \]

### Step 2: Expand the General Term
Expanding this gives:

\[ T_{r+1} = \binom{9}{r} (2^{9-r} x^{9-r}) y^r = \binom{9}{r} 2^{9-r} x^{9-r} y^r \]

### Step 3: Set Up the Condition for Coefficients
We want the term where \( x^{9-r} = x^6 \) and \( y^r = y^3 \). This implies:

1. \( 9 - r = 6 \) (for \( x \))
2. \( r = 3 \) (for \( y \))

### Step 4: Solve for \( r \)
From \( 9 - r = 6 \):

\[ r = 3 \]

### Step 5: Substitute \( r \) into the General Term
Now, we substitute \( r = 3 \) into the general term:

\[ T_{4} = \binom{9}{3} (2x)^{9-3} y^3 = \binom{9}{3} (2x)^6 y^3 \]

### Step 6: Calculate the Coefficient
Now we calculate the coefficient of \( x^6 y^3 \):

\[ T_{4} = \binom{9}{3} (2^6 x^6) y^3 \]

The coefficient is:

\[ \binom{9}{3} \cdot 2^6 \]

### Step 7: Calculate \( \binom{9}{3} \) and \( 2^6 \)
Calculating \( \binom{9}{3} \):

\[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]

Calculating \( 2^6 \):

\[ 2^6 = 64 \]

### Step 8: Final Coefficient Calculation
Now, multiply these values to find the coefficient:

\[ \text{Coefficient} = 84 \times 64 = 5376 \]

Thus, the coefficient of \( x^6 y^3 \) in the expansion of \( (2x + y)^9 \) is **5376**.

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