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JEE Mains · Maths · STD 11 - 7. binomial theoram

ધારોકે \(\left(x-\frac{3}{x^2}\right)^n, x \neq 0 . n \in N\) ના વિસ્તરણમાં પ્રથમ ત્રણ પદોના સહગુણકોનો સરવાળો \(376\) છે. તો \(x^4\) નો સહગુણક \(..........\) છે.

  1. A \(404\)
  2. B \(403\)
  3. C \(402\)
  4. D \(405\)
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Answer & Solution

Correct Answer

(D) \(405\)

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Detailed explanation

Given Binomial \(\left(x-\frac{3}{x^2}\right)^n, x \neq 0, n \in N\) Sum of coefficients of first three terms \({ }^n C_0-{ }^n C_1 \cdot 3+{ }^n C_2 3^2=376\) \(\Rightarrow 3 n^2-5 n-250=0\) \(\Rightarrow(n-10)(3 n+25)=0\) \(\Rightarrow n =10\) Now general term…
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