$In the expansion of (x^{3} + 2 x^{2} + x + 4)^{15}, the coefficient of x^{2} is not divisible by$

8

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25

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27

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64

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Solution

The correct option is C $27$
$(x^{3} + 2 x^{2} + x + 4)^{15} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{45} x^{45} Differentiating w.r.t. x, we get 15 (x^{3} + 2 x^{2} + x + 4)^{14} (3 x^{2} + 4 x + 1) = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} \dots + 45 a_{45} x^{44} Differentiating again w.r.t. x, we get 15 [14 (x^{3} + 2 x^{2} + x + 4)^{13} (3 x^{2} + 4 x + 1)^{2} + (x^{3} + 2 x^{2} + x + 4)^{14} (6 x + 4)] = 2 a_{2} + 6 a_{3} x \dots + 1980 a_{45} x^{43} Putting x=0, we get 2 a_{2} = 15 \times 4^{13} \times (14 + 16) \Rightarrow a_{2} = 2^{26} \times 3^{2} \times 5^{2} So, a_{2} is not divisible by 27.$

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Similar questions

Q. Coefficient of

x^{2}

in the expansion of

(x^{3} + 2 x^{2} + x + 4)^{15}

is-

12 x^{2} - 4 \sqrt{15} x + 5 = 0

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Solve the following systems of equations:

$\frac{10}{x + y} + \frac{2}{x - y} = 4$

$\frac{15}{x + y} - \frac{9}{x - y} = - 2$

Q. Solve for x and y:

\frac{10}{x + y} + \frac{2}{x - y} = 4 \frac{15}{x + y} - \frac{9}{x - y} = - 2

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In the expansion of (x3+2x2+x+4)15, the coefficient of x2 is not divisible by

$In the expansion of (x^{3} + 2 x^{2} + x + 4)^{15}, the coefficient of x^{2} is not divisible by$