To determine the coefficient of $x^6$ in the expansion of $(x^2 + 3x)^{12}$, we use the binomial theorem. The binomial theorem states that:

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

Here:

• $a = x^2$
• $b = 3x$
• $n = 12$

The general term in the expansion is:

$T_k = \binom{12}{k} (x^2)^{12-k} (3x)^k$

Simplify the terms:

• $(x^2)^{12-k} = x^{2(12-k)} = x^{24-2k}$
• $(3x)^k = 3^k x^k$

Thus, the general term becomes:

$T_k = \binom{12}{k} \cdot 3^k \cdot x^{24-2k} \cdot x^k = \binom{12}{k} \cdot 3^k \cdot x^{24-2k+k} = \binom{12}{k} \cdot 3^k \cdot x^{24-k}$

We need the power of $x$ to be 6: $24 - k = 6$ $k = 18$

Coefficient Calculation

The coefficient of $x^6$ corresponds to $T_{18}$. Substitute $k = 18$ into $T_k$:

$T_{18} = \binom{12}{18} \cdot 3^{18}$