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7.Binomial Theorem
hard
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
$33$
(JEE MAIN-2023)
Solution
$T _{ k +1}={ }^{ n } C _{ k }( x )^{\frac{ n – k }{2}}(-6)^{ k }( x )^{\frac{-3}{2} k }$
$\frac{ n – k }{2}-\frac{3}{2} k =0$
$n -4 k =0$
$(-5)^{ n }-\left({ }^{ n } C _{\frac{ n }{}}(-6)^{\frac{ n }{4}}\right)=649$
By observation $(625+24=649)$, we get $n=4$
$\because n=4 \quad k=1$
Required is coefficient of $x^{-4}$ is $\left(\sqrt{4}-\frac{6}{x^{\frac{3}{2}}}\right)^4$ ${ }^4(-6)^3$
By calculating we will get $\lambda=36$
Standard 11
Mathematics
