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

\({\left( {{x^2} + \frac{2}{x}} \right)^{15}}\) ના વિસ્તરણમાં \(x^{15}\) ના સહગુણક અને અચળ પદનો ગુણોત્તર મેળવો.

  1. A \(7: 16\)
  2. B \(7:64\)
  3. C \(1: 4\)
  4. D \(1: 32\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(1: 32\)

Step-by-step Solution

Detailed explanation

\(T_{r+1}=^{15} C_{r}\left(x^{2}\right)^{15-r} \cdot\left(2 x^{-1}\right)^r\) \(=15 \mathrm{C}_{r} \times(2)^{r} \times x^{30-3r}\) For independent term, \(30-3 r=0\) \(\Rightarrow r=10\) Hence the term independent of \(x\).…
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