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(D) The given binomial expression is $\left(x-\frac{3}{x^2}\right)^n$. The first three terms in the expansion are given by the binomial theorem as: $T

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$405$

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(D) The given binomial expression is $\left(x-\frac{3}{x^2}ight)^n$. The first three terms in the expansion are given by the binomial theorem as: $T_1 = { }^n C_0 (x)^n = { }^n C_0 x^n$ $T_2 = { }^n C_1 (x)^{n-1} (-\frac{3}{x^2}) = -3 { }^n C_1 x^{n-3}$ $T_3 = { }^n C_2 (x)^{n-2} (-\frac{3}{x^2})^2 = 9 { }^n C_2 x^{n-6}$ The sum of the coefficients is given as $376$: ${ }^n C_0 - 3 { }^n C_1 + 9 { }^n C_2 = 376$ $1 - 3n + 9 \frac{n(n-1)}{2} = 376$ $2 - 6n + 9n^2 - 9n = 752$ $9n^2 - 15n - 750 = 0$ Dividing by $3$: $3n^2 - 5n - 250 = 0$ $(n-10)(3n+25) = 0$ Since $n \in N$,we have $n = 10$. The general term is $T_{r+1} = { }^{10} C_r (x)^{10-r} (-\frac{3}{x^2})^r = { }^{10} C_r (-3)^r x^{10-3r}$. To find the coefficient of $x^4$,we set $10-3r = 4$,which gives $3r = 6$,so $r = 2$. The coefficient is ${ }^{10} C_2 (-3)^2 = 45 	imes 9 = 405$.

Let the sum of the coefficients of the first three terms in the expansion of $\left(x-\frac{3}{x^2}\right)^n, x \neq 0, n \in N$,be $376$. Then the coefficient of $x^4$ is $......$

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