The value of $n$ for which the expression $x^{4} + 4 x^{3} + n x^{2} + 4 x + 1$ becomes a perfect square is :

3

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4

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5

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6

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Solution

The correct option is D $6$
$(x + 1)^{4} = x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1$
$∴$ The value of $n$ is $6$ .

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