ExamBro
ExamBro
JEE Mains · Maths · STD 11 - 7. binomial theoram

If the constant term in the expansion of \(\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}, x \neq 0\), is \(\alpha \times 2^8 \times \sqrt[5]{3}\), then \(25 \alpha\) is equal to :

  1. A \(639\)
  2. B \(724\)
  3. C \(693\)
  4. D \(742\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(693\)

Step-by-step Solution

Detailed explanation

\( \mathrm{T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}}\left(\frac{3^{1 / 5}}{\mathrm{x}}\right)^{12-\mathrm{r}}\left(\frac{2 \mathrm{x}}{5^{1 / 3}}\right)^{\mathrm{r}} \)…
Same subject
Explore more questions on app