Courses
Courses for Kids
Free study material
Offline Centres
More

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

Last updated date: 26th Feb 2024
Total views: 359.7k
Views today: 9.59k
Verified
359.7k+ views
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.