The coefficient of `x^5` in the expansion of `(x+3)^8`

Answer

To find the coefficient of \( x^5 \) in the expansion of \( (x + 3)^8 \), we will use the Binomial Theorem, which states that:

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

In our case, \( a = x \), \( b = 3 \), and \( n = 8 \).

### Step 1: Identify the terms in the expansion
We need to find the term that contains \( x^5 \). In the general term of the expansion, the \( k \)-th term is given by:

\[ T_k = \binom{n}{k} a^{n-k} b^k = \binom{8}{k} x^{8-k} (3)^k \]

### Step 2: Set up the equation for \( x^5 \)
To find the coefficient of \( x^5 \), we need \( 8 - k = 5 \). This gives us:

\[ k = 8 - 5 = 3 \]

### Step 3: Substitute \( k \) into the term
Now, substitute \( k = 3 \) into the general term:

\[ T_3 = \binom{8}{3} x^{8-3} (3)^3 = \binom{8}{3} x^5 (3)^3 \]

### Step 4: Calculate \( \binom{8}{3} \) and \( 3^3 \)
First, calculate \( \binom{8}{3} \):

\[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56 \]

Next, calculate \( 3^3 \):

\[ 3^3 = 27 \]

### Step 5: Find the coefficient
Now, we can find the coefficient of \( x^5 \):

\[ \text{Coefficient of } x^5 = \binom{8}{3} \cdot 3^3 = 56 \cdot 27 \]

Calculating \( 56 \cdot 27 \):

\[ 56 \cdot 27 = 1512 \]

### Final Answer
Thus, the coefficient of \( x^5 \) in the expansion of \( (x + 3)^8 \) is **1512**.

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