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 $\frac{a}{b}$ is equal to

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

Answer 1

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

$t_{r+1}={}^n{C}_r{x}^{n-r}{y}^r\left(\right(x+y{)}^n)$

$\therefore t_5={}^n{C}_4{x}^{n-4}{y}^4$

$\therefore t_6={}^n{C}_5{x}^{n-5}{y}^r$

$\therefore t_5+t_6=0$

$\therefore {}^n{C}_4{x}^{n-4}{y}^4+{}^n{C}_5{x}^{n-5}{y}^5=0$

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

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

$\therefore \frac{1}{(n-4)!4!}+\frac{1}{(n-5)!5!}\frac{b}{a}=0$

$\therefore \frac{1}{(n-4)}+\frac{1}{5}\frac{b}{a}=0$

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

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

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