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+1 vote
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in Binomial Theorem by (3.3k points)
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How many terms are free from radical signs in the expansion of \((x^{1/5} + y^{1/10})\)55 .

1 Answer

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Best answer

The general term in the expansion of \((x^\frac{1}{5}+y^\frac{1}{10})^{55}\) is given by 

Tr + 1 = 55Cr \((x^\frac{1}{5})^{55-r}\) \((y^\frac{1}{10})^{r}\)

Tr + 155Cr x11-r/5 \(y^\frac{r}{10}\)

Clearly, Tr + 1 will be free from radical signs, if \(\frac{r}{5}\) and \(\frac{r}{10}\)are integers for 0 ≤ r ≤ 55. 

∴ r = 0, 10, 20, 30, 40, 50.

Hence, there are 6 terms in the expansion of \((x^\frac{1}{5}+y^\frac{1}{10})^{55}\) which are independent of radical sign.

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