The coefficient of `x^(5)` in the expansion of `(x +3)^(6)`,is

Options

A. 18
B. 6
C. 12
D. 10

Answer

To find the coefficient of \( x^5 \) in the expansion of \( (x + 3)^6 \), we can use the Binomial Theorem. The Binomial Theorem states that:

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

In our case, \( a = x \), \( b = 3 \), and \( n = 6 \). The general term \( T_r \) in the expansion can be expressed as:

\[ T_r = \binom{n}{r} a^{n-r} b^r = \binom{6}{r} x^{6-r} (3)^r \]

We want the term where the power of \( x \) is 5, which means we need to find \( r \) such that:

\[ 6 - r = 5 \implies r = 1 \]

Now we can substitute \( r = 1 \) into the general term:

\[ T_1 = \binom{6}{1} x^{6-1} (3)^1 \]

Calculating \( T_1 \):

1. Calculate \( \binom{6}{1} \):
\[ \binom{6}{1} = 6 \]

2. Substitute \( r = 1 \) into the term:
\[ T_1 = 6 \cdot x^5 \cdot 3 \]

3. Simplify:
\[ T_1 = 18 x^5 \]

Thus, the coefficient of \( x^5 \) in the expansion of \( (x + 3)^6 \) is \( 18 \).

### Final Answer:
The coefficient of \( x^5 \) is \( 18 \).