Let α be the constant term in the binomial expansion of (√x- 6 x^ 3 2)^n,n≤ 15. If the sum of the coefficients of the remaining terms in the…

Given,

$${(\sqrt{x}-\frac{6}{{x}^{\frac{3}{2}}})}^{n},n\leq 15$$(x​−x23​6​)n,n≤15

Now ${r}^{th}$rth term is given by

$${T}_{r+1}=Crn{({x}^{\frac{1}{2}})}^{n-r}{(-6{x}^{\frac{-3}{2}})}^{r}$$Tr+1​=Crn(x21​)n−r(−6x2−3​)r

$$\Rightarrow {T}_{r+1}={(-6)}^{r}\cdot Crn\cdot {x}^{\frac{n-r}{2}}\cdot {x}^{\frac{-3r}{2}}$$⇒Tr+1​=(−6)r⋅Crn⋅x2n−r​⋅x2−3r​

$$\Rightarrow {T}_{r+1}={(-6)}^{r}\cdot Crn\cdot {x}^{\frac{n-4r}{2}}$$⇒Tr+1​=(−6)r⋅Crn⋅x2n−4r​

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

So, $\alpha =C\frac{n}{4}n{(-6)}^{\frac{n}{4}}$α=C4n​n(−6)4n​

And sum of all coefficients will be ${(\sqrt{1}-\frac{6}{{1}^{\frac{3}{2}}})}^{n}={(-5)}^{n}$(1​−123​6​)n=(−5)n

Also given, ${(-5)}^{n}-C\frac{n}{4}n{(-6)}^{\frac{n}{4}}=649$(−5)n−C4n​n(−6)4n​=649

Now by hit and trial we get, $n=4$n=4

Hence, $\alpha =C14{(-6)}^{\frac{4}{4}}=-24$α=C14(−6)44​=−24

Now for coefficient of ${x}^{-4}$x−4

$$\frac{n-4r}{2}=-4$$2n−4r​=−4

$$\Rightarrow n=4r-8\Rightarrow r=3$$⇒n=4r−8⇒r=3

So, $\lambda (\alpha )={(-6)}^{3}\cdot C34$λ(α)=(−6)3⋅C34

$$\Rightarrow \lambda (-24)={(-6)}^{3}\cdot C34$$⇒λ(−24)=(−6)3⋅C34

$$\Rightarrow \lambda =36$$⇒λ=36