Question

Find $n,$ if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of ${\left( {\sqrt[4]{2} + \dfrac{1}{{\sqrt[4]{3}}}} \right)^n}$ is $\sqrt 6 :1$.

Answer 1

In the given question, we have been asked to find the value of ‘n’ and it is given that the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of ${\left( {\sqrt[4]{2} + \dfrac{1}{{\sqrt[4]{3}}}} \right)^n}$ is $\sqrt 6 :1$ . To solve this question, we need to get ‘n’ on one side of the “equals” sign, and all the other numbers on the other side. To solve this equation for a given variable ‘n’, we have to undo the mathematical operations such as addition, subtraction, multiplication, and division that have been done to the variables.

Complete step by step solution:
We can write ${T_5} = {T_{4 + 1}}$. Now, we have ‘r’ is equals to four, ‘a’ is equals to $\sqrt[4]{2}$ and ‘b’ is equal to $\left( {\dfrac{1}{{\sqrt[4]{3}}}} \right)$. Now, we know that fifth term from the beginning is given as,

\Rightarrow{T_5} = {}^n{C_4}{({(2)^{\dfrac{1}{4}}})^{n - 4}}{\left( {\dfrac{1}{{{{(3)}^{\dfrac{1}{4}}}}}} \right)^4} \\\ $$ $$\Rightarrow{T_5} = {}^n{C_4}{(2)^{\dfrac{1}{4} \times (n - 4)}}\left( {\dfrac{1}{3}} \right)$$ Now, we know that fifth term from the end is given as, $ \Rightarrow {(n - 3)^{th}}$ term from the beginning, Now, we know that ${(n - 3)^{th}}$ term from the beginning is given as, $${T_{n - 4 + 1}} = {}^n{C_{n - 4}}{(\sqrt[4]{2})^{n - (n - 4)}}{\left( {\dfrac{1}{{\sqrt[4]{3}}}} \right)^{n - 4}} \\\ \Rightarrow{T_{n - 3}} = {}^n{C_{n - 4}}(2){\left( {\dfrac{1}{{(3)}}} \right)^{\dfrac{{n - 4}}{4}}} \\\ $$ Now, we have given that the ratio of the fifth term from the beginning to the fifth term from the end is , $\dfrac{{{T_5}}}{{{T_{n - 3}}}} = \dfrac{{\sqrt 6 }}{1}$ $$\Rightarrow\dfrac{{{}^n{C_4}{{(2)}^{\dfrac{1}{4} \times (n - 4)}}\left( {\dfrac{1}{3}} \right)}}{{{}^n{C_{n - 4}}(2){{\left( {\dfrac{1}{{(3)}}} \right)}^{\dfrac{{n - 4}}{4}}}}} = \dfrac{{\sqrt 6 }}{1}$$ $$\Rightarrow\dfrac{{{}^n{C_4}}}{{{}^n{C_{n - 4}}}} \times {(2)^{\dfrac{{n - 8}}{4}}} \times {\left( {\dfrac{1}{{(3)}}} \right)^{\dfrac{{ - (n - 8)}}{4}}} = \dfrac{{\sqrt 6 }}{1}$$ $$\Rightarrow\dfrac{{{}^n{C_4}}}{{{}^n{C_{n - 4}}}} \times {(2 \times 3)^{\dfrac{{n - 8}}{4}}} = \dfrac{{\sqrt 6 }}{1} \\\ \Rightarrow\dfrac{{{}^n{C_4}}}{{{}^n{C_{n - 4}}}} \times {(6)^{\dfrac{{n - 8}}{4}}} = \dfrac{{\sqrt 6 }}{1} \\\ $$ Now, compare the exponent, $\dfrac{{n - 8}}{4} = \dfrac{1}{2} \\\ \Rightarrow n - 8 = 2 \\\ \therefore n = 10 \\\ $ **Hence, we get the value of $n = 10$.** **Note:** Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily. When we have the same base , we can equate the exponents. It is the type of question where only mathematical operations such as addition, subtraction, multiplication and division is used.