Question

In the binomial expansion of ${(a-b)}^n$, $n\ge 5$, the sum of the $5th$and $6th$ terms is $0$. Then $\frac{a}{b}=?$

• A. $\frac{(n–5)}{6}$
• B. $\frac{(n–4)}{5}$
• C. $\frac{5}{(n–4)}$
• D. $\frac{6}{(n–5)}$

Answer 1

The correct option is B

$\frac{(n–4)}{5}$

Explanation For The Correct Option:

Finding $\frac{a}{b}$in the given binomial expansion:

$(r+1)th$term in the expansion of ${(a-b)}^n$

$T_{r+1}={}^n{C}_r{a}^{n-r}\left(b^r\right)....\left(i\right)$

Given that $T_5=0\&T_6=0$

$$\begin{gathered} \Rightarrow T_5+T_6=0 \\ \Rightarrow T_{4+1}+T_{5+1}=0 \\ \Rightarrow {}^n{C}_4{a}^{n-4}{\left(-b\right)}^4-{}^n{C}_5{a}^{n-5}{\left(-b\right)}^5=0[\text{from equation}(i\left)\right] \\ \Rightarrow {}^n{C}_4{a}^{n-4}{\left(-b\right)}^4={}^n{C}_5{a}^{n-5}{\left(-b\right)}^5 \\ \Rightarrow \frac{n!}{4!(n-4)!}\frac{a^n}{a^4}{b}^4=\frac{n!}{5!(n-5)!}\frac{a^n}{a^5}{b}^5 \\ \Rightarrow \frac{1}{4!(n-4)(n-5)!}=\frac{1}{5\times 4!(n-5)!}\frac{b}{a} \\ \Rightarrow \frac{1}{(n-4)}=\frac{1}{5}\frac{b}{a} \\ \Rightarrow \frac{a}{b}=\frac{(n-4)}{5} \end{gathered}$$

Hence, the correct answer is option (B).