Question

The number of rational terms in the expansion of ${{\left( \sqrt[4]{5}+\sqrt[5]{4} \right)}^{100}}$ is
(a) 50
(b) 5
(c) 6
(d) 51

Answer 1

Firstly, we have to write the expansion of ${{\left( \sqrt[4]{5}+\sqrt[5]{4} \right)}^{100}}$ using the formula of expansion of ${{\left( a+b \right)}^{n}}$ which is given by ${{T}_{n+1}}{{=}^{n}}{{C}_{r}}{{\left( a \right)}^{n-r}}{{b}^{r}}$ . The rational terms in this expansion can be obtained only when the powers of a and b are integers. We have to equate the powers of a and b to integers, say p and q respectively. On solving the resultant equation, we will get the values of these exponents. We have to look for the number of common terms in both of the exponents which will be the required result.

Complete step by step answer:
We have to find the number of rational terms in the expansion of ${{\left( \sqrt[4]{5}+\sqrt[5]{4} \right)}^{100}}$ . We can see that the given expression is of the form ${{\left( a+b \right)}^{n}}$ . We know that the general term of the expansion of ${{\left( a+b \right)}^{n}}$ is given by
${{T}_{n+1}}{{=}^{n}}{{C}_{r}}{{\left( a \right)}^{n-r}}{{b}^{r}}...\left( i \right)$
Let us compare the given expression with the general form. We can see that $a=\sqrt[4]{5}={{5}^{\dfrac{1}{4}}},b=\sqrt[5]{4}={{4}^{\dfrac{1}{5}}}\text{ and }n=100$ . Let us substitute these values in the formula (i).
$\Rightarrow {{T}_{n+1}}{{=}^{100}}{{C}_{r}}{{\left( 5 \right)}^{\dfrac{1}{4}\left( 100-r \right)}}{{4}^{\dfrac{1}{5}r}}$
Let us simplify the exponents.
$\begin{aligned} & \Rightarrow {{T}_{n+1}}{{=}^{100}}{{C}_{r}}{{\left( 5 \right)}^{\left( \dfrac{100}{4}-\dfrac{r}{4} \right)}}{{4}^{\dfrac{r}{5}}} \\\ & \Rightarrow {{T}_{n+1}}{{=}^{100}}{{C}_{r}}{{\left( 5 \right)}^{\left( 25-\dfrac{r}{4} \right)}}{{4}^{\dfrac{r}{5}}} \\\ \end{aligned}$
We know that for a number to be rational, its exponents must be an integer. Therefore, the exponents of 5 and 4 must be integers.
Let us consider $25-\dfrac{r}{4}=p$ , where p is an integer. Let us simplify this equation.

& \Rightarrow \dfrac{25\times 4}{4}-\dfrac{r}{4}=p \\\ & \Rightarrow \dfrac{100-r}{4}=p \\\ & \Rightarrow 100-r=4p \\\ \end{aligned}$$ The above equation means that $100-r$ is equal to the multiples of 4 (since we substitute different values for p, that is, $p=0,1,2,3,...$ ). $\Rightarrow 100-r=0,4,8,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,88,92,96,100...\left( ii \right)$ Now, let us consider $\dfrac{r}{5}=q$ . $\Rightarrow r=5q$ Therefore, r will be the multiples of 5. $\Rightarrow r=0,5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100...\left( iii \right)$ The given expression will be rational only when both the terms 5 and 4 are rational. We know that both 5 and 4 will be rational for the common terms in (ii) and (iii). Therefore, the rational terms are 0, 20, 40, 60, 80, 100. Thus, the number of rational terms is 6. **So, the correct answer is “Option c”.** **Note:** Students have a chance of making mistakes when writing the formula for expansion of ${{\left( a+b \right)}^{n}}$ and must thoroughly learn the formula to avoid those mistakes. We have considered the multiples only till 100 because $n=100$ . Using the result of the number of rational numbers, we can find the number of irrational numbers. The total number of terms in the expansion of ${{\left( a+b \right)}^{n}}$ will be $n+1$ . Therefore, we can find the number of irrational terms by subtracting the number of rational terms from the total number of terms.