Let the sum of the coefficient of the first three terms in the expansion of (x−3x2)n,x≠0,n∈N , be 376. Then the coefficient of x4 is____

Given binomial expansion is (x−3x2)nSum of the coefficient of first 3 terms nC0− nC1(3)+ nC2⋅32=372 ⇒1−3n+n(n−1)1−2⋅9=372 ⇒9n2−15n−750=0 ⇒(n−10)(9n+75)=0 ∴ n=10Now T2+1= 10C2x8(−3x2)2= 10C2×9×x4 Coefficient x4 in (x−3x2)= 10C2×9=405

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

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answer is 405.

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Detailed Solution

Given binomial expansion is ${(x - \frac{3}{x^{2}})}^{n}$
Sum of the coefficient of first 3 terms
$^{n} C_{0} -^{n} C_{1} (3) +^{n} C_{2} \cdot 3^{2} = 372 \Rightarrow 1 - 3 n + \frac{n (n - 1)}{1 - 2} \cdot 9 = 372 \Rightarrow 9 n^{2} - 15 n - 750 = 0 \Rightarrow (n - 10) (9 n + 75) = 0 ∴ n = 10$
Now $T_{2 + 1} =^{10} C_{2} x^{8} {(- \frac{3}{x^{2}})}^{2} =^{10} C_{2} \times 9 \times x^{4}$
Coefficient $x^{4}$ in $(x - \frac{3}{x^{2}}) =^{10} C_{2} \times 9 = 405$