Coefficient of $x^{2}$ in the expansion of $(x^{3} + 2 x^{2} + x + 4)^{15}$ is-

Prime

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Perfect square

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0

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Composite

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Solution

The correct option is B Perfect square
Let $(x^{3} + 2 x^{2} + x + 4)^{15} = a_{0} + a_{1} x + a_{2} x^{2} + . . .$
Differentiating w.r.t. $x$
$15 (x^{3} + 2 x^{2} + x + 4)^{14} (3 x^{2} + 4 x + 1) = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + . . . . . . .$
Again different w.r.t $x$
$15 {(x^{3} + 2 x^{2} + x + 4)^{14} (6 x + 4) + 14 (x^{3} + 2 x^{2} + x + 4)^{13} (3 x^{2} + 4 x + 1)^{2}} = 2 a_{2} + 6 a_{3} x + . . . . . . . .$
Now put $x = 0$
$\Rightarrow 15 ({4.4}^{14} + {14.4}^{13}) = 2 a_{2}$
$\Rightarrow 2 a_{2} = 15 \times 4^{13} (16 + 14)$
$\Rightarrow a_{2} = (15)^{2} \times 2^{26}$
Which is a perfect square

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Coefficient of $x^{2}$ in the expansion of $(x^{3} + 2 x^{2} + x + 4)^{15}$ is-