JEE Mains · Maths · STD 11 - 7. binomial theoram
If the coefficients of \(x^{7}\) and \(x^{8}\) in the expansion of \(\left(2+\frac{x}{3}\right)^{n}\) are equal, then the value of \(n\) is equal to \(.....\)
- A \(44\)
- B \(55\)
- C \(48\)
- D \(61\)
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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