Question

The last term in the binomial expansion of ${\left( {\sqrt[3]{2} - \dfrac{1}{{\sqrt 2 }}} \right)^n}$ is ${\left( {\dfrac{1}{{3.\sqrt[3]{9}}}} \right)^{{{\log }_3}8}}$. Find the ${5^{th}}$ term from the beginning. Choose the correct option.
(A)\[{}^{10}{C_6}\]
(B)\[\dfrac{1}{2}.{}^{10}{C_6}\]
(C) \[10.{}^{10}{C_6}\]
(D)None of these

Answer 1

Hint: Here we will simply give information by using binomial expansion formula, and n and after that by using nth term formula of binomial expression we will find the required term.

Complete step-by-step answer:
We can write the given term like ${\left( {{2^{\dfrac{1}{3}}} - \dfrac{1}{{\sqrt 2 }}} \right)^n}$
So let the any general term of r is
${T_{n + 1}} = {}^n{c_r}{\left( {{2^{\dfrac{1}{3}}}} \right)^{n - n}}{\left( { - \dfrac{1}{{\sqrt 2 }}} \right)^n} = \dfrac{{{{\left( { - 1} \right)}^n}}}{{{2^{\dfrac{n}{2}}}}} - - - - (1)$
Now simplify the next term which is given us in the question ${\left( {\dfrac{1}{{3.\sqrt[3]{9}}}} \right)^{{{\log }_3}8}}$, we get
$ = {\left( {\dfrac{1}{{{3^{\dfrac{5}{3}}}}}} \right)^{{{\log }_3}8}}$
Now we know some properties of logarithm i.e.
$ \Rightarrow {a^{{{\log }_b}c}} = {c^{{{\log }_b}c}}$
From this property we can see that a and c are interchanges
$ \Rightarrow {\left( {\dfrac{1}{{{3^{\dfrac{5}{3}}}}}} \right)^{{{\log }_3}8}} = {8^{{{\log }_3}{{\left( 3 \right)}^{ - \dfrac{5}{3}}}}} = {8^{ - \dfrac{5}{3}}} = {\left( {{2^3}} \right)^{ - \dfrac{5}{3}}} = {2^{ - 5}}$
Now from the question we have given that ${T_{n + 1}}$ is equal to ${\left( {\dfrac{1}{{{3^{\dfrac{5}{3}}}}}} \right)^{{{\log }_3}8}}$
Now Put the value of both in equation (1), we get
 $ \Rightarrow {\left( { - 1} \right)^n}{2^{ - \dfrac{n}{2}}} = {2^{ - 5}}$
Now we can write the above equation as,
$ \Rightarrow {\left( { - 1} \right)^n}{2^{\dfrac{n}{2}}} = {\left( { - 1} \right)^{10}}{2^{\dfrac{{10}}{2}}}$
Now we know that when base is equal to same then their power is also same to each other
So, $n = 10$
So from the question we have say that we have to find the ${5^{th}}$term from beginning i.e.${T_5}$
Now as we know that $r = 4$, so the value of ${T_5}$ from the equation (1) is
$ \Rightarrow {T_{r + 1}} = {T_5} = {}^{10}{c_4}{\left( {{2^{\dfrac{1}{3}}}} \right)^6}{\left( { - \dfrac{1}{{\sqrt 2 }}} \right)^4}$
Now we apply the concept of combination in the equation we get,
$ \Rightarrow \dfrac{{7 \times 8 \times 9 \times 10}}{{24}} \times {\left( 2 \right)^2} \times \dfrac{1}{{{2^2}}}$
After solving the equation we get
$ \Rightarrow 210$
So the value of ${T_5} = 210$.

Hence the correct option is A.

Note: In these types of questions we should remember the general form of the equation using the combination concept. And also remember that $r = n - 1$, here n is the number of terms which we have to find. As the term given in the question we should do the conversion into log very carefully.