Let be the constant term in the binomial expansion of (x-6x^(3)/(2))^n,n 15.

Question

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

Answer: 36

Accepted Answer

Given expression $(\sqrt{x}-\frac{6}{{x}^{3/2}}{)}^{n}$. Here, $a=\sqrt{x}$ and $b=-\frac{6}{{x}^{3/2}}$.

The $r$-th term of the binomial expansion of $(a+b{)}^{n}$ is given by

${T}_{r}={}^{n}{C}_{r}{a}^{n-r}{b}^{r}$.

Substitute $a$ and $b$ in this formula, we get:

${T}_{r}={}^{n}{C}_{r}(\sqrt{x}{)}^{n-r}{(-\frac{6}{{x}^{3/2}})}^{r}={}^{n}{C}_{r}(-6{)}^{r}{x}^{\frac{n-4r}{2}}$.

The constant term in the binomial expansion is obtained when the power of $x$ in the terms equals zero.

This happens when $\frac{n-4r}{2}=0$, which gives $n=4r$.

$\begin{gathered} {}^{n}{C}_{\frac{n}{4}}(-6{)}^{\frac{n}{4}}=\alpha \ (-5{)}^{n}-{}^{n}{C}_{\frac{n}{4}}(-6{)}^{n/4}=649 \ 	herefore \alpha =-24 \end{gathered}$

Now, for coefficient of ${x}^{-4}$

$\begin{gathered} \frac{n-4r}{2}=-4 \ n=4r-8\Rightarrow r=3 \ \lambda (-24)=(-6{)}^{3}\cdot {}^{4}{C}_{3} \ \Rightarrow \lambda =36 \end{gathered}$

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