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  • Let  be the constant term in the binomial expansion of <img alt="\mathrm{\left ( \sqrt{x}-\frac{6}{x^{\frac{3

Let \alpha be the constant term in the binomial expansion of \mathrm{\left ( \sqrt{x}-\frac{6}{x^{\frac{3}{2}}} \right )^{n}},\mathrm{n\leq 15}. If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of \mathrm{x^{n}} is \mathrm{\lambda \alpha }, then \mathrm{\lambda } is equal to ________.

Option: 1

36


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \mathrm{T}_{\mathrm{k}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{k}}(\mathrm{x})^{\frac{\mathrm{n}-\mathrm{k}}{2}}(-6)^{\mathrm{k}}(\mathrm{x})^{\frac{-3}{2} \mathrm{k}} \\ \end{aligned}

\begin{aligned} & \frac{\mathrm{n}-\mathrm{k}}{2}-\frac{3}{2} \mathrm{k}=0 \\ \end{aligned}

\begin{aligned} & \mathrm{n}-4 k=0 \\ \end{aligned}

\begin{aligned} & (-5)^{\mathrm{n}}-\left({ }_n C_{\frac{\mathrm{n}}{4}}(-6)^{\frac{\mathrm{n}}{4}}\right)=649 \end{aligned}

By observation (625+24=649), we get \mathrm{n}=4

\because \mathrm{n}=4 \: \& \: \mathrm{k}=1

Required is coefficient of

\begin{aligned} & x^{-4} \text { is }\left(\sqrt{4}-\frac{6}{x^{\frac{3}{2}}}\right)^4 \\ & { }^4 C_1(-6)^3 \end{aligned}

By calculating we will get \lambda=36

Posted by

rishi.raj

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