Answer
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Hint: To solve this question, we will first of all calculate the total number of terms in binomial expansion of ${{\left( {{7}^{\dfrac{1}{5}}}-{{3}^{\dfrac{1}{10}}} \right)}^{60}}$ using the formula.
When, ${{\left( a+b \right)}^{n}}$ is expanded then the number of terms is (n + 1). Then, we will calculate the general terms by using formula \[{{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}\] where our expansion is ${{\left( a+b \right)}^{n}}$
Finally, we will form cases for the total number of rational terms then subtract it to the total number of terms to get a result.
Complete step-by-step answer:
We are given the expression as ${{\left( {{7}^{\dfrac{1}{5}}}-{{3}^{\dfrac{1}{10}}} \right)}^{60}}$ for a term given as ${{\left( a+b \right)}^{n}}$ where the general term is given by ${{T}_{r+1}}$ the total number of terms is (n + 1) where n is the power of (a + b).
Also, the general term of binomial expansion of ${{\left( a+b \right)}^{n}}$ is written as
\[{{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}\]
Using this formula of general term of a binomial expansion and the total number of terms as stated above, we get:
The total number of expansion of \[{{\left( {{7}^{\dfrac{1}{5}}}-{{3}^{\dfrac{1}{10}}} \right)}^{60}}\] is \[60+1=61\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
And the general term in binomial expansion of \[{{\left( {{7}^{\dfrac{1}{5}}}-{{3}^{\dfrac{1}{10}}} \right)}^{60}}\] is given as:
\[{{T}_{r+1}}={}^{60}{{C}_{r}}{{7}^{\dfrac{60-r}{5}}}{{\left( -3 \right)}^{\dfrac{r}{10}}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}\]
This is obtained by using $a={{7}^{\dfrac{1}{5}}}\text{ and b=}{{\left( -3 \right)}^{\dfrac{1}{10}}}$ in the above given formula.
Now, to calculate the total number of irrational numbers or terms we will first calculate the total number of rational terms.
Let us first define a rational number:
A number is called rational if it can be written in the form of $\dfrac{p}{q}$ where $q\ne 0$
Observing the term of equation (ii) we observe that ${}^{60}{{C}_{r}}$ is always rational for any r. The point of issue is ${{7}^{\dfrac{60-r}{5}}}\text{ and }{{\left( -3 \right)}^{\dfrac{r}{10}}}$
Consider power of 7 and -3 as $\dfrac{60-r}{5}\text{ and }\dfrac{r}{10}$ respectively then, we will consider cases for possible values of r.
Case I: r = 0, then \[\dfrac{60-r}{5}=\dfrac{60-0}{5}=\dfrac{60}{5}=12\] then ${{7}^{12}}$ is rational.
When r = 0 then $\dfrac{r}{10}=0\text{ and }{{\left( -3 \right)}^{0}}$ is rational.
And the product of two rationals is rational. So (ii) is rational.
Case II: r = 10, then \[\dfrac{60-r}{5}=\dfrac{60-10}{5}=\dfrac{50}{5}=10\] then ${{7}^{10}}$ is rational.
Also when r = 10 then \[\dfrac{r}{10}=\dfrac{10}{10}=1\] and ${{\left( -3 \right)}^{1}}$ is rational.
Again, the product of rational is rational. So the term (ii) is rational.
Case III: r = 20, then \[\dfrac{60-r}{5}=\dfrac{60-20}{5}=\dfrac{40}{5}=8\] then ${{7}^{8}}$ is rational.
Also when r = 20 then \[\dfrac{r}{10}=\dfrac{20}{10}=2\] and ${{\left( -3 \right)}^{2}}$ is rational.
And the product of rational is rational. So, terms in equation (ii) are rational.
Case IV: r = 30, then \[\dfrac{60-r}{5}=\dfrac{60-30}{5}=\dfrac{30}{5}=6\] then ${{7}^{6}}$ is rational.
Also when r = 30 then \[\dfrac{r}{10}=\dfrac{30}{10}=3\] and ${{\left( -3 \right)}^{3}}$ is rational.
Again, the product of two or more rationals is rational. So, the term in equation (ii) is rational.
Case V: r = 40 then \[\dfrac{60-r}{5}=\dfrac{60-40}{5}=\dfrac{20}{5}=4\] then ${{7}^{4}}$ is rational.
Also when r = 40 then \[\dfrac{r}{10}=\dfrac{40}{10}=4\] and ${{\left( -3 \right)}^{4}}$ is rational.
Again, the product of rational is rational. So, equation (ii) is rational.
Case VI: r = 50 then \[\dfrac{60-r}{5}=\dfrac{60-50}{5}=\dfrac{10}{5}=2\] then ${{7}^{2}}$ is rational.
Also when r = 50 then \[\dfrac{r}{10}=\dfrac{50}{10}=5\] and ${{\left( -3 \right)}^{5}}$ is rational.
Again, the product of rational is rational. So, equation (ii) term is rational.
Case VII: r = 60 then \[\dfrac{60-r}{5}=\dfrac{60-60}{5}=\dfrac{0}{5}=0\] then ${{7}^{0}}=1$ is rational
When r = 60 then \[\dfrac{r}{10}=\dfrac{60}{10}=6\] and ${{\left( -3 \right)}^{6}}$ is rational.
Again, the product of rational is rational. So, equation (ii) is rational.
Here, from all above cases when r = 0, 10, 20, 30, 40, 50, 60 then
\[{{T}_{r+1}}={}^{60}{{C}_{r}}{{7}^{\dfrac{60-r}{5}}}{{\left( -3 \right)}^{\dfrac{r}{10}}}\text{ is rational}\text{.}\]
So, we have number of irrational term is = total number of terms of binomial expansion of \[{{\left( {{7}^{\dfrac{1}{5}}}-{{3}^{\dfrac{1}{10}}} \right)}^{60}}\]
Subject to the number of rational terms we have 7 rational terms as we had 7 cases. Also, number of terms = 61 by equation (i)
Then, \[\text{Number of irrational terms }=\text{ 61}-\text{7 }=\text{ 54}\]
Therefore, we have the total number of irrational terms in binomial expansion of ${{\left( {{7}^{\dfrac{1}{5}}}-{{3}^{\dfrac{1}{10}}} \right)}^{60}}$ is 54
So, the correct answer is “Option D”.
Note: Another way to solve this question can be directly calculating the total number of rational terms in binomial expansion of ${{\left( {{7}^{\dfrac{1}{5}}}-{{3}^{\dfrac{1}{10}}} \right)}^{60}}$ by using the fact that, any value of r which is divisible by both 5 and 10 gives rational term in expansion of ${{\left( {{7}^{\dfrac{1}{5}}}-{{3}^{\dfrac{1}{10}}} \right)}^{60}}$
This is so because we had general term as:
\[{{T}_{r+1}}={}^{60}{{C}_{r}}{{7}^{\dfrac{60-r}{5}}}{{\left( -3 \right)}^{\dfrac{r}{10}}}\]
Where powers of 7 and -3 has $\dfrac{60-r}{5}\text{ and }\dfrac{r}{10}$ and we want to make this $\dfrac{60-r}{5}\text{ and }\dfrac{r}{10}$ as an integer. So, any value of r which is divisible by both 5 and 10 then, any term divisible by 10 works as any number divisible by 10 is divisible by 5. So, possibilities of r = 0, 10, 20, 30, 40, 50, 60 for the general term to be rational.
When, ${{\left( a+b \right)}^{n}}$ is expanded then the number of terms is (n + 1). Then, we will calculate the general terms by using formula \[{{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}\] where our expansion is ${{\left( a+b \right)}^{n}}$
Finally, we will form cases for the total number of rational terms then subtract it to the total number of terms to get a result.
Complete step-by-step answer:
We are given the expression as ${{\left( {{7}^{\dfrac{1}{5}}}-{{3}^{\dfrac{1}{10}}} \right)}^{60}}$ for a term given as ${{\left( a+b \right)}^{n}}$ where the general term is given by ${{T}_{r+1}}$ the total number of terms is (n + 1) where n is the power of (a + b).
Also, the general term of binomial expansion of ${{\left( a+b \right)}^{n}}$ is written as
\[{{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}\]
Using this formula of general term of a binomial expansion and the total number of terms as stated above, we get:
The total number of expansion of \[{{\left( {{7}^{\dfrac{1}{5}}}-{{3}^{\dfrac{1}{10}}} \right)}^{60}}\] is \[60+1=61\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
And the general term in binomial expansion of \[{{\left( {{7}^{\dfrac{1}{5}}}-{{3}^{\dfrac{1}{10}}} \right)}^{60}}\] is given as:
\[{{T}_{r+1}}={}^{60}{{C}_{r}}{{7}^{\dfrac{60-r}{5}}}{{\left( -3 \right)}^{\dfrac{r}{10}}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}\]
This is obtained by using $a={{7}^{\dfrac{1}{5}}}\text{ and b=}{{\left( -3 \right)}^{\dfrac{1}{10}}}$ in the above given formula.
Now, to calculate the total number of irrational numbers or terms we will first calculate the total number of rational terms.
Let us first define a rational number:
A number is called rational if it can be written in the form of $\dfrac{p}{q}$ where $q\ne 0$
Observing the term of equation (ii) we observe that ${}^{60}{{C}_{r}}$ is always rational for any r. The point of issue is ${{7}^{\dfrac{60-r}{5}}}\text{ and }{{\left( -3 \right)}^{\dfrac{r}{10}}}$
Consider power of 7 and -3 as $\dfrac{60-r}{5}\text{ and }\dfrac{r}{10}$ respectively then, we will consider cases for possible values of r.
Case I: r = 0, then \[\dfrac{60-r}{5}=\dfrac{60-0}{5}=\dfrac{60}{5}=12\] then ${{7}^{12}}$ is rational.
When r = 0 then $\dfrac{r}{10}=0\text{ and }{{\left( -3 \right)}^{0}}$ is rational.
And the product of two rationals is rational. So (ii) is rational.
Case II: r = 10, then \[\dfrac{60-r}{5}=\dfrac{60-10}{5}=\dfrac{50}{5}=10\] then ${{7}^{10}}$ is rational.
Also when r = 10 then \[\dfrac{r}{10}=\dfrac{10}{10}=1\] and ${{\left( -3 \right)}^{1}}$ is rational.
Again, the product of rational is rational. So the term (ii) is rational.
Case III: r = 20, then \[\dfrac{60-r}{5}=\dfrac{60-20}{5}=\dfrac{40}{5}=8\] then ${{7}^{8}}$ is rational.
Also when r = 20 then \[\dfrac{r}{10}=\dfrac{20}{10}=2\] and ${{\left( -3 \right)}^{2}}$ is rational.
And the product of rational is rational. So, terms in equation (ii) are rational.
Case IV: r = 30, then \[\dfrac{60-r}{5}=\dfrac{60-30}{5}=\dfrac{30}{5}=6\] then ${{7}^{6}}$ is rational.
Also when r = 30 then \[\dfrac{r}{10}=\dfrac{30}{10}=3\] and ${{\left( -3 \right)}^{3}}$ is rational.
Again, the product of two or more rationals is rational. So, the term in equation (ii) is rational.
Case V: r = 40 then \[\dfrac{60-r}{5}=\dfrac{60-40}{5}=\dfrac{20}{5}=4\] then ${{7}^{4}}$ is rational.
Also when r = 40 then \[\dfrac{r}{10}=\dfrac{40}{10}=4\] and ${{\left( -3 \right)}^{4}}$ is rational.
Again, the product of rational is rational. So, equation (ii) is rational.
Case VI: r = 50 then \[\dfrac{60-r}{5}=\dfrac{60-50}{5}=\dfrac{10}{5}=2\] then ${{7}^{2}}$ is rational.
Also when r = 50 then \[\dfrac{r}{10}=\dfrac{50}{10}=5\] and ${{\left( -3 \right)}^{5}}$ is rational.
Again, the product of rational is rational. So, equation (ii) term is rational.
Case VII: r = 60 then \[\dfrac{60-r}{5}=\dfrac{60-60}{5}=\dfrac{0}{5}=0\] then ${{7}^{0}}=1$ is rational
When r = 60 then \[\dfrac{r}{10}=\dfrac{60}{10}=6\] and ${{\left( -3 \right)}^{6}}$ is rational.
Again, the product of rational is rational. So, equation (ii) is rational.
Here, from all above cases when r = 0, 10, 20, 30, 40, 50, 60 then
\[{{T}_{r+1}}={}^{60}{{C}_{r}}{{7}^{\dfrac{60-r}{5}}}{{\left( -3 \right)}^{\dfrac{r}{10}}}\text{ is rational}\text{.}\]
So, we have number of irrational term is = total number of terms of binomial expansion of \[{{\left( {{7}^{\dfrac{1}{5}}}-{{3}^{\dfrac{1}{10}}} \right)}^{60}}\]
Subject to the number of rational terms we have 7 rational terms as we had 7 cases. Also, number of terms = 61 by equation (i)
Then, \[\text{Number of irrational terms }=\text{ 61}-\text{7 }=\text{ 54}\]
Therefore, we have the total number of irrational terms in binomial expansion of ${{\left( {{7}^{\dfrac{1}{5}}}-{{3}^{\dfrac{1}{10}}} \right)}^{60}}$ is 54
So, the correct answer is “Option D”.
Note: Another way to solve this question can be directly calculating the total number of rational terms in binomial expansion of ${{\left( {{7}^{\dfrac{1}{5}}}-{{3}^{\dfrac{1}{10}}} \right)}^{60}}$ by using the fact that, any value of r which is divisible by both 5 and 10 gives rational term in expansion of ${{\left( {{7}^{\dfrac{1}{5}}}-{{3}^{\dfrac{1}{10}}} \right)}^{60}}$
This is so because we had general term as:
\[{{T}_{r+1}}={}^{60}{{C}_{r}}{{7}^{\dfrac{60-r}{5}}}{{\left( -3 \right)}^{\dfrac{r}{10}}}\]
Where powers of 7 and -3 has $\dfrac{60-r}{5}\text{ and }\dfrac{r}{10}$ and we want to make this $\dfrac{60-r}{5}\text{ and }\dfrac{r}{10}$ as an integer. So, any value of r which is divisible by both 5 and 10 then, any term divisible by 10 works as any number divisible by 10 is divisible by 5. So, possibilities of r = 0, 10, 20, 30, 40, 50, 60 for the general term to be rational.
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