Question

If the last term in the binomial expansion of ${{\left( {{2}^{\dfrac{1}{3}}}-\dfrac{1}{\sqrt{2}} \right)}^{n}}$is ${{\left( \dfrac{1}{{{\left( 3 \right)}^{\dfrac{5}{3}}}} \right)}^{{{\log }_{3}}8}}$then 5th term from the beginning is:
a) 210
b) 420
c) 105
d) None of these

Answer 1

Hint: Use the formula for general term of the binomial expansion of ${{\left( x+y \right)}^{n}}$ is ${}^{n}{{C}_{r}}.{{x}^{n-r}}.{{y}^{r}}$. From the last term we can find the value of n then we can find the 5th term from the beginning.

Complete step-by-step answer:


The formula for general term of the binomial expansion of${{\left( x+y \right)}^{n}}$is:
${{T}_{r+1}}={}^{n}{{C}_{r}}.{{x}^{n-r}}.{{y}^{r}}$

In the above expression, r can take value from 0, 1, 2 …n and r + 1 is the nth term.

${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$

The general term for the binomial expansion of ${{\left( {{2}^{\dfrac{1}{3}}}-\dfrac{1}{\sqrt{2}} \right)}^{n}}$is:

${{T}_{r+1}}={}^{n}{{C}_{r}}.{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{n-r}}.{{\left( -\dfrac{1}{\sqrt{2}} \right)}^{r}}$

For the finding the last term, put r = n in the above equation.

${{T}_{n+1}}={}^{n}{{C}_{n}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{n-n}}{{\left( -\dfrac{1}{\sqrt{2}} \right)}^{n}}$

The value of ${}^{n}{{C}_{n}}$ in the above expression is 1 so we can write the above expression as:

 ${{T}_{n+1}}=1{{\left( -\dfrac{1}{\sqrt{2}} \right)}^{n}}$………….. Eq. (1)

For finding the value of n, we need to equate the above expression to${{\left( \dfrac{1}{{{\left( 3 \right)}^{\dfrac{5}{3}}}} \right)}^{{{\log }_{3}}8}}$.

The value of ${{\left( \dfrac{1}{{{\left( 3 \right)}^{\dfrac{5}{3}}}} \right)}^{{{\log }_{3}}8}}$is solved below:

$\begin{align}

  & {{\left( {{3}^{-\dfrac{5}{3}}} \right)}^{{{\log }_{3}}8}} \\

 & {{\left( 3 \right)}^{-\dfrac{5}{3}{{\log }_{3}}8}} \\

 & {{\left( 3 \right)}^{{{\log }_{3}}{{8}^{-\dfrac{5}{3}}}}} \\

 & {{8}^{-\dfrac{5}{3}}} \\

\end{align}$

The properties of logarithm that we have used above in the calculation are:

$\begin{align}

  & n{{\log }_{b}}a \\

 & = {{\log }_{b}}{{a}^{n}} \\

\end{align}$

And another property of logarithm that we have used is:

${{a}^{{{\log }_{a}}b}}=b$

Now, equating ${{8}^{-\dfrac{5}{3}}}$to eq. (1) we get,

$\begin{align}

  & {{\left( -\dfrac{1}{\sqrt{2}} \right)}^{n}}={{8}^{-\dfrac{5}{3}}} \\

 & {{\left( -{{2}^{-\dfrac{1}{2}}} \right)}^{n}}={{8}^{-\dfrac{5}{3}}} \\

 & {{\left( -1 \right)}^{n}}{{\left( 2 \right)}^{-\dfrac{n}{2}}}={{2}^{-5}} \\

 & n=10 \\

\end{align}$

We have got the value of n as 10.

Now, we are going to calculate the 5th term using the general formula of the nth term and in which we are substituting the value of r as 4 and n =10.

$\begin{align}

  & {{T}_{5}}={}^{10}{{C}_{4}}{{\left( 2 \right)}^{\dfrac{6}{3}}}{{\left( -\dfrac{1}{\sqrt{2}} \right)}^{4}} \\

 & \Rightarrow {{T}_{5}}={}^{10}{{C}_{4}}{{\left( 2 \right)}^{2}}\left( \dfrac{1}{4} \right) \\

 & \Rightarrow {{T}_{5}}=210 \\

\end{align}$

Hence, the 5th term from the beginning is 210.

Hence, the correct option is (a).

Note: There is a property of ${}^{n}{{C}_{r}}$ that n and r cannot be negative.