Question

A ratio of the fifth term from the beginning to the $5^{th}$ term from the end in binomial expansion \[{\left( {{2^{1/3}} + \dfrac{1}{{2{{(3)}^{1/3}}}}} \right)^{10}}\]

Answer 1

Hint: We use the concept of binomial expansion and count the value of r when starting from beginning and when starting from end. Write the terms and find their ratio by dividing one term by another.
* A binomial expansion helps us to expand expressions of the form \[{(a + b)^n}\]through the formula \[{(a + b)^n} = \sum\limits_{r = 0}^n {^n{C_r}{{(a)}^{n - r}}{{(b)}^r}} \]
* Formula of combination is given by\[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], where factorial is expanded by the formula \[n! = n \times (n - 1)! = n \times (n - 1) \times (n - 2)!.... = n \times (n - 1) \times (n - 2)....3 \times 2 \times 1\]
* Ratio of any number ‘x’ to ‘y’ is given by \[x:y = \dfrac{x}{y}\]

Complete step-by-step solution:
We are given the term\[{\left( {{2^{1/3}} + \dfrac{1}{{2{{(3)}^{1/3}}}}} \right)^{10}}\] ……….… (1)
Here \[n = 10;a = {2^{1/3}};b = \dfrac{1}{{2{{(3)}^{1/3}}}}\]
We use binomial expansion to expand the given term
\[ \Rightarrow {\left( {{2^{1/3}} + \dfrac{1}{{2{{(3)}^{1/3}}}}} \right)^{10}} = \sum\limits_{r = 0}^{10} {^{10}{C_r}{{\left( {{2^{1/3}}} \right)}^{10 - r}}{{\left( {\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)}^r}} \]
\[ \Rightarrow {\left( {{2^{1/3}} + \dfrac{1}{{2{{(3)}^{1/3}}}}} \right)^{10}}{ = ^{10}}{C_0}{\left( {{2^{1/3}}} \right)^{10 - 0}}{\left( {\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)^0}{ + ^{10}}{C_1}{\left( {{2^{1/3}}} \right)^{10 - 1}}{\left( {\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)^1} + ......{ + ^{10}}{C_{10}}{\left( {{2^{1/3}}} \right)^{10 - 10}}{\left( {\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)^{10}}\]
From this expansion we can write the fifth term from starting has \[r = 4\] and the fifth term from end has \[r = 6\].
We find the terms separately and then find the ratio.
Fifth term from starting:
Here \[n = 10;a = {2^{1/3}};b = \dfrac{1}{{2{{(3)}^{1/3}}}};r = 4\]
Let the fifth term from end be denoted by ‘x’
\[\Rightarrow x{ = ^{10}}{C_4}{\left( {{2^{1/3}}} \right)^{10 - 4}}{\left( {\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)^4}\]
\[ \Rightarrow x{ = ^{10}}{C_4}{\left( {{2^{1/3}}} \right)^6}{\left( {\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)^4}\]
Use combination formula\[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], where factorial is expanded by the formula \[n! = n \times (n - 1)! = n \times (n - 1) \times (n - 2)!.... = n \times (n - 1) \times (n - 2)....3 \times 2 \times 1\]
\[\Rightarrow x = \dfrac{{10!}}{{6!4!}}{\left( {{2^{1/3}}} \right)^6}{\left( {\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)^4}\]..................… (1)
Fifth term from ending:
Here \[n = 10;a = {2^{1/3}};b = \dfrac{1}{{2{{(3)}^{1/3}}}};r = 6\]
Let the fifth term from end be denoted by ‘y’
\[\Rightarrow y{ = ^{10}}{C_6}{\left( {{2^{1/3}}} \right)^{10 - 6}}{\left( {\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)^6}\]
\[ \Rightarrow y{ = ^{10}}{C_6}{\left( {{2^{1/3}}} \right)^4}{\left( {\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)^6}\]
Use combination formula\[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], where factorial is expanded by the formula \[n! = n \times (n - 1)! = n \times (n - 1) \times (n - 2)!.... = n \times (n - 1) \times (n - 2)....3 \times 2 \times 1\]
\[\Rightarrow y = \dfrac{{10!}}{{6!4!}}{\left( {{2^{1/3}}} \right)^4}{\left( {\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)^6}\]............… (2)
Now we find the ratio of two terms by dividing equation (1) by (2)
\[ \Rightarrow \dfrac{x}{y} = \dfrac{{\dfrac{{10!}}{{6!4!}}{{\left( {{2^{1/3}}} \right)}^6}{{\left( {\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)}^4}}}{{\dfrac{{10!}}{{6!4!}}{{\left( {{2^{1/3}}} \right)}^4}{{\left( {\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)}^6}}}\]
Cancel same terms from numerator and denominator
\[\Rightarrow \dfrac{x}{y} = \dfrac{{{{\left( {{2^{1/3}}} \right)}^6}{{\left( {\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)}^4}}}{{{{\left( {{2^{1/3}}} \right)}^4}{{\left( {\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)}^6}}}\]
We use the formula \[{a^6} = {a^{4 + 2}} = {a^4}.{a^2}\]to expand terms in numerator and denominator
\[\Rightarrow \dfrac{x}{y} = \dfrac{{{{\left( {{2^{1/3}}} \right)}^4}{{\left( {{2^{1/3}}} \right)}^2}{{\left({\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)}^4}}}{{{{\left( {{2^{1/3}}} \right)}^4}{{\left( {\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)}^4}{{\left( {\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)}^2}}}\]
Cancel same terms from numerator and denominator
\[\Rightarrow \dfrac{x}{y} = \dfrac{{{{\left( {{2^{1/3}}} \right)}^2}}}{{{{\left( {\dfrac{1}{{2{{(3)}^{1/3}}}}} \right)}^2}}}\]
Solve the denominator using \[{\left( {{a^m}} \right)^n} = {a^{mn}}\]
\[ \Rightarrow \dfrac{x}{y} = \dfrac{{{2^{2/3}}}}{{\dfrac{1}{{{2^2}{{(3)}^{2/3}}}}}}\]
Make fraction simpler
\[ \Rightarrow \dfrac{x}{y} = \dfrac{{{2^{2/3}}{2^2}{{(3)}^{2/3}}}}{1}\]
\[ \Rightarrow \dfrac{x}{y} = \dfrac{{4 \times {2^{2/3}}{{(3)}^{2/3}}}}{1}\]
We can write \[{2^{2/3}} = {({2^2})^{1/3}} = {4^{1/3}}\]and\[{3^{2/3}} = {({3^2})^{1/3}} = {9^{1/3}}\]
\[ \Rightarrow \dfrac{x}{y} = \dfrac{{4 \times {4^{1/3}} \times {9^{1/3}}}}{1}\]
Since we know when power is same base can be multiplied
\[ \Rightarrow \dfrac{x}{y} = \dfrac{{4 \times {{(4 \times 9)}^{1/3}}}}{1}\]
\[ \Rightarrow \dfrac{x}{y} = \dfrac{{4 \times {{36}^{1/3}}}}{1}\]
Ratio of \[x:y = 4 \times {36^{1/3}}:1\]

\[\therefore \]Ratio of fifth term from staring to the fifth term from end is \[4 \times {36^{1/3}}:1\]

Note: Students many times make the mistake of writing the fifth term from starting and ending as the same i.e. having \[r = 5\]. This is wrong as students start writing the values of r from 1, we always start writing the value of r from 0 to 10, so the fifth term from starting comes different from fifth term from end.