Mathematics - Binomial Theorem Question with Solution | TestHub

Given,

${\left(\sqrt{x}-\frac{6}{x^{\frac{3}{2}}}\right)}^n,n\le 15$

Now  $r^{th}$ term is given by

$T_{r+1}={}^{n}C_r{\left(x^{\frac{1}{2}}\right)}^{n-r}{\left(-6x^{\frac{-3}{2}}\right)}^r$

$\Rightarrow T_{r+1}={\left(-6\right)}^r·{}^{n}C_r·x^{\frac{n-r}{2}}·x^{\frac{-3r}{2}}$

$\Rightarrow T_{r+1}={\left(-6\right)}^r·{}^{n}C_r·x^{\frac{n-4r}{2}}$

Now we have to take, $n=4r$ for constant term so $n$ is divisible by $4.$

So, $\alpha ={}^{n}C_{\frac{n}{4}}{\left(-6\right)}^{\frac{n}{4}}$

And sum of all coefficients will be ${\left(\sqrt{1}-\frac{6}{1^{\frac{3}{2}}}\right)}^n={\left(-5\right)}^n$

Also given, ${\left(-5\right)}^n-{}^{n}C_{\frac{n}{4}}{\left(-6\right)}^{\frac{n}{4}}=649$

Now by hit and trial we get, $n=4$

Hence, $\alpha ={}^{4}C_1{\left(-6\right)}^{\frac{4}{4}}=-24$

Now for coefficient of $x^{-4}$

$\frac{n-4r}{2}=-4$

$\Rightarrow n=4r-8\Rightarrow \boxed{r=3}$

So, $\lambda \left(\alpha \right)={\left(-6\right)}^3·{}^{4}C_3$

$\Rightarrow \lambda \left(-24\right)={\left(-6\right)}^3·{}^{4}C_3$

$\Rightarrow \lambda =36$