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

\(\left(2+\frac{x}{3}\right)^{n}\) ના વિસ્તરણમાં જો \(x^{7}\) અને \(x^{8}\) ના સહગુણક સમાન હોય તો \(n\) ની કિમંત મેળવો.

  1. A \(44\)
  2. B \(55\)
  3. C \(48\)
  4. D \(61\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(55\)

Step-by-step Solution

Detailed explanation

\({ }^{n} C_{7} 2^{n-7} \frac{1}{3^{7}}=^{n} C_{8} 2^{n-8} \frac{1}{3^{8}}\) \(\Rightarrow \frac{n !}{(n-7) ! 7 !} 2^{n-7} \frac{1}{3^{7}}=\frac{n !}{(n-8) ! 8 !} 2^{n-8} \frac{1}{3^{8}} \Rightarrow \frac{1}{(n-7)}=\frac{1}{8} \cdot \frac{1}{2} \cdot \frac{1}{3}\)…
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