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Question: In the binomial expansion of \( {\left( {a - b} \right)^n} \) , \( n \geqslant 5 \) , the sum of 5th...

In the binomial expansion of (a−b)n{\left( {a - b} \right)^n} , n⩾5n \geqslant 5 , the sum of 5th and 6th terms is zero then a/b equal to
A. 5n−4\dfrac{5}{{n - 4}}
B. 6n−5\dfrac{6}{{n - 5}}
C. n−56\dfrac{{n - 5}}{6}
D. n−45\dfrac{{n - 4}}{5}

Explanation

Solution

Hint : Here we have a limit for n that n must be greater than or equal to 5. So first expand (a−b)n{\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 (x−y)n{\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 (a−b)n{\left( {a - b} \right)^n} , n⩾5n \geqslant 5 , the sum of 5th and 6th terms is zero.
We have to find the ratio of a:b, ab\dfrac{a}{b}
First we are expanding (a−b)n{\left( {a - b} \right)^n} using the binomial expansion of (x−y)n{\left( {x - y} \right)^n} where x is a and y is b.
Therefore, the binomial expansion of (a−b)n{\left( {a - b} \right)^n} is
\mathop \sum \limits_{r = 0}^n {\left( { - 1} \right)^r}.{}_{}^nC_r^{}{a^{n - r}}{b^r}
⇒[(−1)0.nC0an−0b0]+[(−1)1.nC1an−1b1]+[(−1)2.nC2an−2b2]+[(−1)3.nC3an−3b3]+[(−1)4.nC4an−4b4]+[(−1)5.nC5an−5b5]+.....\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] + .....
⇒nC0an−nC1an−1b+nC2an−2b2−nC3an−3b3+nC4an−4b4−nC5an−5b5+.....\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 nC4an−4b4{}_{}^nC_4^{}{a^{n - 4}}{b^4} and 6th term is −nC5an−5b5- {}_{}^nC_5^{}{a^{n - 5}}{b^5} .
The sum of 5th and 6th terms is zero as given in the question.
⇒nC4an−4b4−nC5an−5b5=0\Rightarrow {}_{}^nC_4^{}{a^{n - 4}}{b^4} - {}_{}^nC_5^{}{a^{n - 5}}{b^5} = 0
⇒nC4an−4b4=nC5an−5b5\Rightarrow {}_{}^nC_4^{}{a^{n - 4}}{b^4} = {}_{}^nC_5^{}{a^{n - 5}}{b^5}
⇒nC4nC5=an−5b5an−4b4\Rightarrow \dfrac{{{}_{}^nC_4^{}}}{{{}_{}^nC_5^{}}} = \dfrac{{{a^{n - 5}}{b^5}}}{{{a^{n - 4}}{b^4}}}
⇒(n!4!(n−4)!)(n!5!(n−5)!)=ba\Rightarrow \dfrac{{\left( {\dfrac{{n!}}{{4!\left( {n - 4} \right)!}}} \right)}}{{\left( {\dfrac{{n!}}{{5!\left( {n - 5} \right)!}}} \right)}} = \dfrac{b}{a}
⇒5n−4=ba\Rightarrow \dfrac{5}{{n - 4}} = \dfrac{b}{a}
Invert the numerator and denominator
∴ab=n−45\therefore \dfrac{a}{b} = \dfrac{{n - 4}}{5}
Hence, the correct option is Option D, the value of ab\dfrac{a}{b} is n−45\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 (a−b)n{\left( {a - b} \right)^n} , which is
Tr+1=(−1)r.nCran−rbr{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, T4+1=T5=(−1)4.nC4an−4b4=nC4an−4b4{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, T5+1=T6=(−1)5.nC5an−5b5=−nC5an−5b5{T_{5 + 1}} = {T_6} = {\left( { - 1} \right)^5}.{}_{}^nC_5^{}{a^{n - 5}}{b^5} = - {}_{}^nC_5^{}{a^{n - 5}}{b^5}