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}{{{3^{\dfrac{5}{3}}}}}} \right)^{{{\log }_3}8}} $ , then the 5th term from the beginning is
A. 210
B. 420
C. 105
D. None of these

Answer 1

Hint: Find the last term of the given binomial expansion using the general form of a term formula of binomial expansion which is mentioned below and equate it to the given value and then get the value of n. So with this n value we can get the value of the 5th term.
General term of $ {\left( {x - y} \right)^n} $ is $ {T_{r + 1}} = {\left( { - 1} \right)^r}.{}_{}^nC_r^{}{x^{n - r}}{y^r} $

Complete step-by-step answer:
We are given that the last term in the binomial expansion of $ {\left( {{2^{\dfrac{1}{3}}} - \dfrac{1}{{\sqrt 2 }}} \right)^n} $ is $ {\left( {\dfrac{1}{{{3^{\dfrac{5}{3}}}}}} \right)^{{{\log }_3}8}} $ .
We have to find the value of its 5th term.
Considering $ {2^{\dfrac{1}{3}}} $ as x and $ \dfrac{1}{{\sqrt 2 }} $ as y, $ {\left( {{2^{\dfrac{1}{3}}} - \dfrac{1}{{\sqrt 2 }}} \right)^n} $ becomes $ {\left( {x - y} \right)^n} $
So the general term of $ {\left( {x - y} \right)^n} $ is $ {T_{r + 1}} = {\left( { - 1} \right)^r}.{}_{}^nC_r^{}{x^{n - r}}{y^r} $
For the last term of an expansion, the value of r is n as $ n + 1 $ is the last term.
Therefore, the last term is $ {T_{n + 1}} = {\left( { - 1} \right)^n}.{}_{}^nC_n^{}{x^{n - n}}{y^n} $
So the last term of $ {\left( {{2^{\dfrac{1}{3}}} - \dfrac{1}{{\sqrt 2 }}} \right)^n} $ is $ {T_{n + 1}} = {\left( { - 1} \right)^n}.{}_{}^nC_n^{}{\left( {{2^{\dfrac{1}{3}}}} \right)^{n - n}}{\left( {\dfrac{1}{{\sqrt 2 }}} \right)^n} $
The value of $ {}_{}^nC_n^{} $ is 1 and anything power zero is 1.
 $ \Rightarrow {T_{n + 1}} = {\left( { - 1} \right)^n}{\left( {\dfrac{1}{{\sqrt 2 }}} \right)^n} $
 $ \Rightarrow {T_{n + 1}} = {\left( {\dfrac{{ - 1}}{{\sqrt 2 }}} \right)^n} $
We are given that the value of the last term is $ {\left( {\dfrac{1}{{{3^{\dfrac{5}{3}}}}}} \right)^{{{\log }_3}8}} $ , 8 is the cube of 2, which can also be written as $ {\left( {\dfrac{1}{{{3^{\dfrac{5}{3}}}}}} \right)^{3{{\log }_3}2}} $ as [$ \log {a^m} = m\log a $]
 $ \Rightarrow {3^{\dfrac{{ - 5}}{3} \times 3{{\log }_3}2}} = {3^{ - 5{{\log }_3}2}} $
 $ \Rightarrow {3^{{{\log }_3}{2^{ - 5}}}} $ as [$ m\log a = \log {a^m} $]
 $ \Rightarrow {3^{{{\log }_3}{2^{ - 5}}}} = {2^{ - 5}} $ as [$ {a^{{{\log }_a}b}} = b $ ]
The last term of the expansion is $ {2^{ - 5}} $ which is equal to $ {\left( {\dfrac{{ - 1}}{{\sqrt 2 }}} \right)^n} $
 $ {\left( {\dfrac{{ - 1}}{{\sqrt 2 }}} \right)^n} = {2^{ - 5}} $
 $ \Rightarrow \dfrac{{{{\left( { - 1} \right)}^n}}}{{{2^{\left( {\dfrac{n}{2}} \right)}}}} = {2^{ - 5}} $
 $ \Rightarrow \dfrac{{{{\left( { - 1} \right)}^n}}}{{{2^{\left( {\dfrac{n}{2}} \right)}}}} = \dfrac{1}{{{2^5}}} $
Numerator is positive so n must be an even number. The bases of the denominators are equal so we are equating their powers.
 $ \Rightarrow \dfrac{n}{2} = 5 \Rightarrow n = 2 \times 5 = 10 $
The value of n is 10.
For the 5th term of the expansion, the value of r is 4, n is 10.
  $ {T_{4 + 1}} = {\left( { - 1} \right)^4}.{}_{}^{10}C_4^{}{\left( {{2^{\dfrac{1}{3}}}} \right)^{10 - 4}}{\left( {\dfrac{1}{{\sqrt 2 }}} \right)^4} $
 $ \Rightarrow {T_5} = {}_{}^{10}C_4^{}{\left( {{2^{\dfrac{1}{3}}}} \right)^6}{\left( {\dfrac{1}{{\sqrt 2 }}} \right)^4} $ as even powers of -1 is 1.
The value of $ {}_{}^{10}C_4^{} $ is 210
 $ \Rightarrow {T_5} = 210 \times \left( {{2^{\dfrac{6}{3}}}} \right) \times \dfrac{1}{{{{\left( {\sqrt 2 } \right)}^4}}} $
 $ \Rightarrow {T_5} = 210 \times \left( {{2^2}} \right) \times \dfrac{1}{{{2^2}}} = 210 $
Therefore, the 5th term of $ {\left( {{2^{\dfrac{1}{3}}} - \dfrac{1}{{\sqrt 2 }}} \right)^{10}} $ is $210$.
So, the correct answer is “Option A”.

Note: For every binomial expansion with degree n, we will have $ n + 1 $ terms. So $ n + 1 $ th term will be the last term for every expansion. Be careful with the exponents as when we send a negative exponent from numerator to denominator it becomes positive and vice-versa. Odd powers of -1 give -1 itself.