Question

In the binomial expansion of $ {\left( {a - b} \right)^n} $ , $ n \geqslant 5 $ , the sum of 5th and 6th terms is zero then a/b equal to
A. $ \dfrac{5}{{n - 4}} $
B. $ \dfrac{6}{{n - 5}} $
C. $ \dfrac{{n - 5}}{6} $
D. $ \dfrac{{n - 4}}{5} $

Answer 1

Hint: Here we have a limit for n that n must be greater than or equal to 5. So first expand $ {\left( {a - b} \right)^n} $ using the below mentioned binomial theorem formula by putting x as a and y as b. From the binomial expansion take out the 5th and 6th terms; add them and equate it to zero. Then find the division of a over b.
Formula used:
Binomial expansion of $ {\left( {x - y} \right)^n} $ is $ \mathop \sum \limits_{r = 0}^n {\left( { - 1} \right)^r}.{}_{}^nC_r^{}{x^{n - r}}{y^r} $ , where ‘r’ must be less than or equal to 1.

Complete step-by-step answer:
We are given that in the binomial expansion of $ {\left( {a - b} \right)^n} $ , $ n \geqslant 5 $ , the sum of 5th and 6th terms is zero.
We have to find the ratio of a:b, $ \dfrac{a}{b} $
First we are expanding $ {\left( {a - b} \right)^n} $ using the binomial expansion of $ {\left( {x - y} \right)^n} $ where x is a and y is b.
Therefore, the binomial expansion of $ {\left( {a - b} \right)^n} $ is
$ \mathop \sum \limits_{r = 0}^n {\left( { - 1} \right)^r}.{}_{}^nC_r^{}{a^{n - r}}{b^r} $
$ \Rightarrow \left[ {{{\left( { - 1} \right)}^0}.{}_{}^nC_0^{}{a^{n - 0}}{b^0}} \right] + \left[ {{{\left( { - 1} \right)}^1}.{}_{}^nC_1^{}{a^{n - 1}}{b^1}} \right] + \left[ {{{\left( { - 1} \right)}^2}.{}_{}^nC_2^{}{a^{n - 2}}{b^2}} \right] + \left[ {{{\left( { - 1} \right)}^3}.{}_{}^nC_3^{}{a^{n - 3}}{b^3}} \right] + \left[ {{{\left( { - 1} \right)}^4}.{}_{}^nC_4^{}{a^{n - 4}}{b^4}} \right] + \left[ {{{\left( { - 1} \right)}^5}.{}_{}^nC_5^{}{a^{n - 5}}{b^5}} \right] + ..... $
 $ \Rightarrow {}_{}^nC_0^{}{a^n} - {}_{}^nC_1^{}{a^{n - 1}}b + {}_{}^nC_2^{}{a^{n - 2}}{b^2} - {}_{}^nC_3^{}{a^{n - 3}}{b^3} + {}_{}^nC_4^{}{a^{n - 4}}{b^4} - {}_{}^nC_5^{}{a^{n - 5}}{b^5} + ..... $
As we can see in the above expansion 5th term is $ {}_{}^nC_4^{}{a^{n - 4}}{b^4} $ and 6th term is $ - {}_{}^nC_5^{}{a^{n - 5}}{b^5} $ .
The sum of 5th and 6th terms is zero as given in the question.
 $ \Rightarrow {}_{}^nC_4^{}{a^{n - 4}}{b^4} - {}_{}^nC_5^{}{a^{n - 5}}{b^5} = 0 $
 $ \Rightarrow {}_{}^nC_4^{}{a^{n - 4}}{b^4} = {}_{}^nC_5^{}{a^{n - 5}}{b^5} $
 $ \Rightarrow \dfrac{{{}_{}^nC_4^{}}}{{{}_{}^nC_5^{}}} = \dfrac{{{a^{n - 5}}{b^5}}}{{{a^{n - 4}}{b^4}}} $
 $ \Rightarrow \dfrac{{\left( {\dfrac{{n!}}{{4!\left( {n - 4} \right)!}}} \right)}}{{\left( {\dfrac{{n!}}{{5!\left( {n - 5} \right)!}}} \right)}} = \dfrac{b}{a} $
 $ \Rightarrow \dfrac{5}{{n - 4}} = \dfrac{b}{a} $
Invert the numerator and denominator
 $ \therefore \dfrac{a}{b} = \dfrac{{n - 4}}{5} $
Hence, the correct option is Option D, the value of $ \dfrac{a}{b} $ is $ \dfrac{{n - 4}}{5} $
So, the correct answer is “Option D”.

Note: Instead of expanding the total expression, we can directly find the 5th and 6th terms using a formula. This formula is the general form of the terms of a binomial expansion of $ {\left( {a - b} \right)^n} $ , which is
 $ {T_{r + 1}} = {\left( { - 1} \right)^r}.{}_{}^nC_r^{}{a^{n - r}}{b^r} $ , where r can be less than or equal to n
Here we need 5th and 6th terms, so just substitute r=4 and r=5 respectively to get the terms.
When r is equal to 4, $ {T_{4 + 1}} = {T_5} = {\left( { - 1} \right)^4}.{}_{}^nC_4^{}{a^{n - 4}}{b^4} = {}_{}^nC_4^{}{a^{n - 4}}{b^4} $
When r is equal to 5, $ {T_{5 + 1}} = {T_6} = {\left( { - 1} \right)^5}.{}_{}^nC_5^{}{a^{n - 5}}{b^5} = - {}_{}^nC_5^{}{a^{n - 5}}{b^5} $