<p>Coefficient of \(x^2\) in the expansion of \((x^3 + 2x^2 + x + 4)^{15}\) is</p>
Step-by-Step Solution
Key Concept: Use the multinomial theorem to identify which products of terms from each factor give x². Since the minimum power of x in the base polynomial is 0 (from the constant 4), we need to track combinations that yield exactly x².
<p><strong>Step 1:</strong> Recognize that $(x^3 + 2x^2 + x + 4)^{15}$ requires multinomial expansion. We need the coefficient of $x^2$ when selecting terms from each of the 15 identical factors.</p><p><strong>Step 2:</strong> To get $x^2$, the possible selections are:</p><ul><li>Select $x$ from 2 factors and $4$ from 13 factors: $\binom{15}{2} \cdot 1^2 \cdot 4^{13}$</li><li>Select $2x^2$ from 1 factor and $4$ from 14 factors: $\binom{15}{1} \cdot 2 \cdot 4^{14}$</li></ul><p><strong>Step 3:</strong> Calculate each contribution:</p><ul><li>From $x^2$ terms: $\binom{15}{2} \cdot 4^{13} = 105 \cdot 4^{13}$</li><li>From $2x^2$ terms: $15 \cdot 2 \cdot 4^{14} = 30 \cdot 4^{14} = 30 \cdot 4 \cdot 4^{13} = 120 \cdot 4^{13}$</li></ul><p><strong>Step 4:</strong> Sum the coefficients: $105 \cdot 4^{13} + 120 \cdot 4^{13} = 225 \cdot 4^{13} = 225 \cdot 2^{26}$</p><p>∴ Answer: <strong>C</strong></p>
Correct Answer: C