Question

find the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of ${{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{10}}$.
(a) $1:4{{\left( 16 \right)}^{\dfrac{1}{3}}}$
(b) $1:2{{\left( 6 \right)}^{\dfrac{1}{3}}}$
(c) $2{{\left( 36 \right)}^{\dfrac{1}{3}}}:1$
(d) $4{{\left( 36 \right)}^{\dfrac{1}{3}}}:1$

Answer 1

In this question, we have to first find the binomial expansion of ${{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{10}}$.
Using the formula of binomial expansion of elements say $a$ and $b$ raised to the power $n$ which is given by ${{\left( a+b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{\left( a \right)}^{n}}{{\left( b \right)}^{0}}{{+}^{n}}{{C}_{1}}{{\left( a \right)}^{n+1}}{{\left( b \right)}^{1}}+...{{+}^{n}}{{C}_{r}}{{\left( a \right)}^{n-r}}{{\left( b \right)}^{r}}+...{{+}^{n}}{{C}_{n-1}}{{\left( a \right)}^{1}}{{\left( b \right)}^{n-1}}{{+}^{n}}{{C}_{n}}{{\left( a \right)}^{0}}{{\left( b \right)}^{n}}$Where we have $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Also since the number of terms in the binomial expansion of ${{\left( a+b \right)}^{n}}$ is equal to $n+1$. Using this we will have that the number of terms in the binomial expansion of ${{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{10}}$ is equals to 11. After finding the binomial expansion of ${{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{10}}$ we will have to determine the fifth term from the beginning to the fifth term from the end in the binomial expansion and then find the ratio of the same.

Complete step by step answer:
Let us first determine the binomial expansion of ${{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{10}}$.
Since we know that the binomial expansion of elements say $a$ and $b$ raised to the power $n$ which is given by ${{\left( a+b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{\left( a \right)}^{n}}{{\left( b \right)}^{0}}{{+}^{n}}{{C}_{1}}{{\left( a \right)}^{n+1}}{{\left( b \right)}^{1}}+...{{+}^{n}}{{C}_{r}}{{\left( a \right)}^{n-r}}{{\left( b \right)}^{r}}+...{{+}^{n}}{{C}_{n-1}}{{\left( a \right)}^{1}}{{\left( b \right)}^{n-1}}{{+}^{n}}{{C}_{n}}{{\left( a \right)}^{0}}{{\left( b \right)}^{n}}$Where we have $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
On comparing the expression ${{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{10}}$ with ${{\left( a+b \right)}^{n}}$, we get that
$a={{2}^{\dfrac{1}{3}}}$, $b=\dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}}$ and $n=10$.
Now since we know that the number of terms in the binomial expansion of ${{\left( a+b \right)}^{n}}$ is equal to $n+1$.
Using this we will have that the number of terms in the binomial expansion of ${{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{10}}$ is equals to
$10+1=11$
Now we will evaluate ${{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{10}}$ using the formula ${{\left( a+b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{\left( a \right)}^{n}}{{\left( b \right)}^{0}}{{+}^{n}}{{C}_{1}}{{\left( a \right)}^{n+1}}{{\left( b \right)}^{1}}+...{{+}^{n}}{{C}_{r}}{{\left( a \right)}^{n-r}}{{\left( b \right)}^{r}}+...{{+}^{n}}{{C}_{n-1}}{{\left( a \right)}^{1}}{{\left( b \right)}^{n-1}}{{+}^{n}}{{C}_{n}}{{\left( a \right)}^{0}}{{\left( b \right)}^{n}}$.
Then we get

& {{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{10}}{{=}^{10}}{{C}_{0}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{10}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{0}}{{+}^{10}}{{C}_{1}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{10-1}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{1}}+... \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{+}^{10}}{{C}_{6}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{10-6}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{6}}+...{{+}^{10}}{{C}_{9}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{1}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{9}}{{+}^{10}}{{C}_{10}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{0}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{10}} \\\ \end{aligned}$$ Now since the first term from the beginning of the above binomial expansion is given by $$^{10}{{C}_{0}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{10}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{0}}$$ Therefore the fifth term from the beginning of the above binomial expansion is given by $$^{10}{{C}_{4}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{10-4}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{4}}{{=}^{10}}{{C}_{4}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{6}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{4}}..........(1)$$ Also the last term of the above binomial expansion is given $$^{10}{{C}_{10}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{0}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{10}}$$ Therefore the fifth term from the end of the above binomial expansion is given by $$^{10}{{C}_{6}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{10-6}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{6}}{{=}^{10}}{{C}_{6}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{4}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{6}}..............(2)$$ Now in order to determine the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of $${{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{10}}$$. We have to divide the expression in equation (1) by the expression in equation (2). Then we get $$\dfrac{^{10}{{C}_{4}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{6}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{4}}}{^{10}{{C}_{6}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{4}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{6}}}..............(3)$$ Now on calculating the value of $$^{10}{{C}_{6}}$$ using $$^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$$ , we get $$^{10}{{C}_{6}}=\dfrac{10!}{6!4!}$$ Also we have $$^{10}{{C}_{4}}=\dfrac{10!}{4!6!}$$ Since the value $$^{10}{{C}_{6}}=\dfrac{10!}{6!4!}{{=}^{10}}{{C}_{4}}$$, therefore from (3) we have $$\dfrac{^{10}{{C}_{6}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{6}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{4}}}{^{10}{{C}_{6}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{4}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{6}}}=\dfrac{{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{6}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{4}}}{{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{4}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{6}}}$$ Now on calculating the above expression we get $$\begin{aligned} & \dfrac{{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{6}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{4}}}{{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{4}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{6}}}={{\left( {{2}^{\dfrac{1}{3}}} \right)}^{6-4}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{4-6}} \\\ & ={{\left( {{2}^{\dfrac{1}{3}}} \right)}^{2}}{{\left( \dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{-2}} \\\ & =\left( {{2}^{\dfrac{2}{3}}} \right)\left( \dfrac{1}{{{2}^{-2}}{{\left( 3 \right)}^{\dfrac{-2}{3}}}} \right) \\\ & =\left( {{2}^{\dfrac{2}{3}}} \right)\left( {{2}^{2}}{{\left( 3 \right)}^{\dfrac{2}{3}}} \right) \\\ & =4\left( {{\left( 2 \right)}^{\dfrac{2}{3}}}{{\left( 3 \right)}^{\dfrac{2}{3}}} \right) \\\ & =4{{\left( 6 \right)}^{\dfrac{2}{3}}} \\\ & =4{{\left( {{6}^{2}} \right)}^{\dfrac{1}{3}}} \\\ & ==4{{\left( 36 \right)}^{\dfrac{1}{3}}} \end{aligned}$$ Hence we have the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of $${{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{10}}$$ is equals to $$4{{\left( 36 \right)}^{\dfrac{1}{3}}}:1$$. **So, the correct answer is “Option D”.** **Note:** In this problem, in order to determine the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of $${{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{10}}$$ We have to find the binomial expansion of $${{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{2{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right)}^{10}}$$. While calculating the fifth term from the beginning we have carefully chosen the term by seeing the first term of the expansion.