Let alpha be the constant term in the binomia | vedclass

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(C) The general term is $T_{k+1} = {^nC_k} (x^{1/2})^{n-k} (-6 x^{-3/2})^k = {^nC_k} (-6)^k x^{(n-4k)/2}$.For the constant term,$n-4k = 0$,so $n = 4k$

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(C) The general term is $T_{k+1} = {^nC_k} (x^{1/2})^{n-k} (-6 x^{-3/2})^k = {^nC_k} (-6)^k x^{(n-4k)/2}$.For the constant term,$n-4k = 0$,so $n = 4k$. Since $n \leq 15$,$k$ can be $1, 2, 3$.The sum of all coefficients is found by setting $x=1$,which is $(1-6)^n = (-5)^n$.The sum of the remaining terms is $(-5)^n - \alpha = 649$.If $k=1, n=4$: $(-5)^4 - {^4C_1}(-6)^1 = 625 + 24 = 649$. This holds true.Thus,$n=4$ and $\alpha = {^4C_1}(-6)^1 = -24$.The coefficient of $x^{-n} = x^{-4}$ occurs when $(n-4k)/2 = -4$,so $4-4k = -8$,$4k = 12$,$k=3$.The coefficient is ${^4C_3}(-6)^3 = 4 	imes (-216) = -864$.We have $-864 = \lambda(-24)$,so $\lambda = \frac{-864}{-24} = 36$.

Let $\alpha$ be the constant term in the binomial expansion of $\left(\sqrt{x} - \frac{6}{x^{3/2}}\right)^n$,$n \leq 15$. 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 $..........$.

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