Question

The number of terms which are free from radical signs in the expansion of ${{\left( {{y}^{\dfrac{1}{5}}}+{{x}^{\dfrac{1}{10}}} \right)}^{55}}$ are:
(a) 5
(b) 6
(c) 7
(d) None of these

Answer 1

Hint: To solve this given question, we will first find out the general term of the above expansion. This general term will be a function of a variable, r. Now, to remove the terms from the radical sign, we will put those values of r such that the power on x and y becomes the whole number. We will count these values of r and that will be the answer.

Complete step by step solution:
To start with, we will first find out the general terms of the above expansion. The general term or the ${{r}^{th}}$ term of an expression ${{\left( a+b \right)}^{p}}$ is given by the formula $^{p}{{C}_{r}}{{\left( a \right)}^{p-r}}{{\left( b \right)}^{r}}.$ In our case, the value of ‘a’ is ${{y}^{\dfrac{1}{5}}}$ and the value of b is ${{x}^{\dfrac{1}{10}}}.$ Thus, the general term in our case will be given by,
$\text{General Term }={{\text{ }}^{55}}{{C}_{r}}{{\left( {{y}^{\dfrac{1}{5}}} \right)}^{55-r}}{{\left( {{x}^{\dfrac{1}{10}}} \right)}^{r}}$
Now, we will use the following exponential identity, ${{\left( {{m}^{a}} \right)}^{b}}={{m}^{a\times b}}$
Thus, we will get,
$\Rightarrow \text{General Term }={{\text{ }}^{55}}{{C}_{r}}\left( {{y}^{\dfrac{1}{5}\times \left( 55-r \right)}} \right)\left( {{x}^{\dfrac{1}{10}\times r}} \right)$
$\Rightarrow \text{General Term }={{\text{ }}^{55}}{{C}_{r}}\left( {{y}^{\dfrac{55-r}{5}}} \right)\left( {{x}^{\dfrac{r}{10}}} \right)$
$\Rightarrow \text{General Term }={{\text{ }}^{55}}{{C}_{r}}\left( {{y}^{\dfrac{55}{5}-\dfrac{r}{5}}} \right)\left( {{x}^{\dfrac{r}{10}}} \right)$
$\Rightarrow \text{General Term }={{\text{ }}^{55}}{{C}_{r}}\left( {{y}^{11-\dfrac{r}{5}}} \right)\left( {{x}^{\dfrac{r}{10}}} \right)$
Now, we have to put these values of r such that the power on x and y are integers. Now, the range of r is $0\le r\le 55.$ Now, we will try to free y from the radical. Thus, $11-\dfrac{r}{5}$ should be an integer. Now, as we know that, $0\le r\le 55$ we will multiply all sides by – 1. Thus, we will get,
$0\ge -r\ge -55$
Now, we will divide all the parts by 5. Thus, we will get,
$\Rightarrow 0\ge -\dfrac{r}{5}\ge \dfrac{-55}{5}$
$\Rightarrow 0\ge -\dfrac{r}{5}\ge -11$
Now, $11-\dfrac{r}{5}$ is an integer. So, $11-\dfrac{r}{5}$ will be equal to 0, 1, 2, 3, 4, 5, …… 10, 11. Therefore, the values of r we will get as,
$11-\dfrac{r}{5}=0$
$\Rightarrow 11=\dfrac{r}{5}$
$\Rightarrow r=55$
$11-\dfrac{r}{5}=1$
$\Rightarrow 10=\dfrac{r}{5}$
$\Rightarrow r=50$
Similarly, we will make other cases also and find other values. Thus the values of r = 0, 5, 10, 15, 20, ….. 50, 55.
Now, we will try to free x from the radical. Thus, $\dfrac{r}{10}$ should be an integer. Now, as we know that, $0\le r\le 55,$ we will divide all the parts by 10. Thus, we will get,
$0\le \dfrac{r}{10}\le \dfrac{55}{10}$
$\Rightarrow 0\le \dfrac{r}{10}\le 5.5$
Thus, $\dfrac{r}{10}$ can have values 0, 1, 2, 3, 4, 5. Therefore, the values of r are 0, 10, 20, 30, 40, 50.
Now, we will take the common values of r as the answer. The common values of r = 0, 10, 20, 30, 40, 50. Thus, there are a total of 6 values of r.
Hence, option (b) is the right answer.

Note: The above question can be solved alternatively in the following way. The general term of the expansion of ${{\left( {{y}^{\dfrac{1}{5}}}+{{x}^{\dfrac{1}{10}}} \right)}^{55}}$ will be:
$\text{General Term }={{\text{ }}^{55}}{{C}_{r}}{{\left( {{y}^{\dfrac{1}{5}}} \right)}^{55-r}}{{\left( {{x}^{\dfrac{1}{10}}} \right)}^{r}}$
$\Rightarrow \text{General Term }={{\text{ }}^{55}}{{C}_{r}}\left( {{y}^{11-\dfrac{r}{5}}} \right)\left( {{x}^{\dfrac{r}{10}}} \right)$
$\Rightarrow \text{General Term }={{\text{ }}^{55}}{{C}_{r}}\left( \dfrac{{{y}^{11}}}{{{y}^{\dfrac{r}{5}}}} \right)\left( {{x}^{\dfrac{r}{10}}} \right)$
$\Rightarrow \text{General Term }={{\text{ }}^{55}}{{C}_{r}}\left( {{y}^{11}} \right)\left( \dfrac{{{x}^{\dfrac{r}{10}}}}{{{y}^{\dfrac{r}{5}}}} \right)$
$\Rightarrow \text{General Term }={{\text{ }}^{55}}{{C}_{r}}\left( {{y}^{11}} \right){{\left( \dfrac{{{x}^{\dfrac{1}{10}}}}{{{y}^{\dfrac{1}{5}}}} \right)}^{r}}$
$\Rightarrow \text{General Term }={{\text{ }}^{55}}{{C}_{r}}\left( {{y}^{11}} \right){{\left( \dfrac{x}{{{y}^{^{2}}}} \right)}^{\dfrac{r}{10}}}$
Now, for the power to be an integer, r should be multiple of 10. Thus, r = 0, 10, 20, 30, 40, 50. Therefore, the total value of r is 6.