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 :
- A \(639\)
- B \(724\)
- C \(693\)
- D \(742\)
Answer & Solution
Correct Answer
(C) \(693\)
Step-by-step Solution
Detailed explanation
\( \mathrm{T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}}\left(\frac{3^{1 / 5}}{\mathrm{x}}\right)^{12-\mathrm{r}}\left(\frac{2 \mathrm{x}}{5^{1 / 3}}\right)^{\mathrm{r}} \)…
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