Question

In the binomial expansion of $(a-b{)}^n,n\ge 5$, the sum of the $5^{th}$ and $6^{th}$ term is zero. Then $A/b$ equals

• A. $(n-5)/6$
• B. $(n-4)/5$
• C. $5/(n-4)$
• D. $6/(n-4)$

Answer 1

The correct option is B $(n-4)/5$

$(a-b{)}^n{\Rightarrow }^n{C}_0{a}^n+..{.}^n{C}_n(-b{)}^n$

$5^{th}term+6^{th}tan=0$

${\therefore }^n{C}_4{a}^{n-4}(-b{)}^4{+}^n{C}_5{a}^{n-5}(-b{)}^5=0$

$\frac{h!}{(n-4)!4!}{a}^{n-4}{b}^4-\frac{n!}{(n-5)!5!}{a}^{n-5}{b}^5=0$

$\frac{n!}{(n-4)(n-5)!4!}\frac{a^n}{a^7}.b^n=\frac{n!}{(n-1)!5\times 4!}\times \frac{a^n}{a^5}{b}^5$

$\frac{1}{n-4}=\frac{1}{5}\frac{b}{a}$

$\frac{a}{b}=\frac{n-4}{5}$

Option B is correct