The number of terms in the expansion of (y^{1 | vedclass

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(A) The general term of the expansion $(y^{1/5} + x^{1/10})^{55}$ is given by:$T_{r+1} = ^{55}C_{r} (y^{1/5})^{55-r} (x^{1/10})^{r}$$T_{r+1} = ^{55}C_

Vedclass · Accepted Answer

$6$

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(A) The general term of the expansion $(y^{1/5} + x^{1/10})^{55}$ is given by:$T_{r+1} = ^{55}C_{r} (y^{1/5})^{55-r} (x^{1/10})^{r}$$T_{r+1} = ^{55}C_{r} y^{\frac{55-r}{5}} x^{\frac{r}{10}}$For the powers of $x$ and $y$ to be free from radical signs,the exponents $\frac{55-r}{5}$ and $\frac{r}{10}$ must be integers.$\frac{55-r}{5} = 11 - \frac{r}{5}$ must be an integer,which implies $r$ must be a multiple of $5$.$\frac{r}{10}$ must be an integer,which implies $r$ must be a multiple of $10$.Combining these,$r$ must be a multiple of $10$ where $0 \le r \le 55$.Possible values for $r$ are $0, 10, 20, 30, 40, 50$.There are $6$ such values of $r$,so there are $6$ terms in the expansion that are free from radical signs.

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

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