Binomial theorem for positive integral Index

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Question

The number of terms which are free from radical signs in the expansion of ${(y^{1 / 5} + x^{1 / 10})}^{55}$ , is

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Solution

The general term in the expansion of ${(y^{1 / 5} + x^{1 / 10})}^{55}$
s given by
$T_{r + 1} =^{55} C_{r} {(y^{1 / 5})}^{55 - r} {(x^{1 / 10})}^{r} =^{55} C_{r} y^{11 - r / 5} x^{r / 10}$
clearly, $T_{r + 1}$ will be independent of radical signs if $r / 5$ and
$r$ / $10$ are integers, where $0 \leq r \leq 55$ .
$r = 0, 10, 20, 30, 40, 50$
Hence, there are $6$ `terms in the expansion of the
given expression which are independent of radicals.

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Binomial theorem for positive integral Index

Remember concepts with our Masterclasses.

The number of terms which are free from radical signs in the expansion of y1/5+x1/1055, is

Ready to Test Your Skills?

free study material

The number of terms which are free from radical signs in the expansion of ${(y^{1 / 5} + x^{1 / 10})}^{55}$ , is