Question

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

• A. $\frac{1}{6}(n-5)$
• B. $\frac{1}{5}(n-4)$
• C. $\frac{5}{n-4}$
• D. $\frac{6}{n-5}$

Answer 1

The correct option is B $\frac{1}{5}(n-4)$

${}^n{C}_4{a}^{n-4}{b}^4={}^n{C}_5{a}^{n-5}{b}^5$
$\frac{n!}{4!(n-4)!}a=\frac{n!}{5!(n-5)!}.b$
Therefore
$\frac{a}{b}=\frac{n-4}{5}$