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

જો \(\left(2 x^{3}+\frac{3}{x}\right)^{10}\) નાં દ્વિપદી વિસ્તરણમાં \(x\) નાં ધન બેકી ધાતવાળા પદોમાંના સહગુણકોનો સરવાળો \(5^{10}-\beta \cdot 3^{9}\) હોય. તો \(\beta\) = ................  

  1. A \(36\)
  2. B \(75\)
  3. C \(89\)
  4. D \(83\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(83\)

Step-by-step Solution

Detailed explanation

\(T_{r+1}={ }^{10} C_{r}\left(2 x^{3}\right)^{10-r}\left(\frac{3}{x}\right)^{r}\) \(={ }^{10} C_{r} 2^{10-r} 3^{r} x^{30-4 r}\) Put \(r=0,1,2, \ldots 7\) and we get \(\beta=83\)
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