JEE Mains · Maths · STD 11 - 7. binomial theoram
Let \(\alpha\) be the constant term in the binomial expansion of \(\left(\sqrt{ x }-\frac{6}{ x ^{\frac{3}{2}}}\right)^{ n }, n \leq 15\). If the sum of the coefficients of the remaining terms in the expansion is \(649\) and the coefficient of \(x^{-n}\) is \(\lambda \alpha\), then \(\lambda\) is equal to \(..........\).
- A \(35\)
- B \(34\)
- C \(36\)
- D \(33\)
Answer & Solution
Correct Answer
(C) \(36\)
Step-by-step Solution
Detailed explanation
\(T _{ k +1}={ }^{ n } C _{ k }( x )^{\frac{ n - k }{2}}(-6)^{ k }( x )^{\frac{-3}{2} k }\) \(\frac{ n - k }{2}-\frac{3}{2} k =0\) \(n -4 k =0\) \((-5)^{ n }-\left({ }^{ n } C _{\frac{ n }{}}(-6)^{\frac{ n }{4}}\right)=649\) By observation \((625+24=649)\), we get \(n=4\)…
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