1f:I[80107,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 20:I[76763,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1e:T442,{"@context":"https://schema.org","@graph":[{"@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https://www.divinejee.com"},{"@type":"ListItem","position":4,"name":"Let alpha be the constant term in the binomial expansion of ","item":"https://www.divinejee.com/question/let-alpha-be-the-constant-term-in-the-binomial-expansion-of-xfrac-6xfrac-32n-n-leq-15-if-the-sum-o"}]},{"@type":"QAPage","mainEntity":{"@type":"Question","name":"Question Let alpha be the constant term in the binomial expansion of on Topic","text":"

Let $\\alpha$ be the constant term in the binomial expansion of $\\left(\\sqrt{x}-\\frac{6}{x^{\\frac{3}{2}}}\\right)^{n}, n \\leq 15$. If the sum of the ...","answerCount":1,"upvoteCount":0,"datePublished":"2026-01-16","author":{"@type":"Organization","name":"DivineJEE"},"acceptedAnswer":{"@type":"Answer","text":"Check the detailed solution on the page.","upvoteCount":0,"url":"https://www.divinejee.com/question/let-alpha-be-the-constant-term-in-the-binomial-expansion-of-xfrac-6xfrac-32n-n-leq-15-if-the-sum-o"}}}]}21:T417,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:

$T_{r} = {}^n{C_r}(\sqrt{x})^{n-r}\left(-\frac{6}{x^{3/2}}\right)^r = {}^n{C_r}(-6)^r x^{\frac{n-4r}{2}}$.

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 $n = 4r$.

$$ \begin{aligned} & { }^n C_{\frac{n}{4}}(-6)^{\frac{n}{4}}=\alpha \\\\ & (-5)^n-{ }^n C_{\frac{n}{4}}(-6)^{n / 4}=649 \\\\ & \text { By observation (625 + 24 = 649) , we get n = 4 } \\\\ & \therefore \alpha=-24 \end{aligned} $$

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

$$ \begin{aligned} & \frac{n-4 r}{2}=-4 \\\\ & n=4 r-8 \Rightarrow r=3 \\\\ & \lambda(-24)=(-6)^3 \cdot{ }^4 C_3 \\\\ & \Rightarrow \lambda=36 \end{aligned} $$