In the binomial expansion of $(a-b)^n$, $n \geq 5$, the sum of $5^{th}$ and $6^{th}$ term is $0$, then $\frac{a}{b}$ is ?
I've solved this problem but its coming $n-4$ only, and the answer says it should be $\displaystyle{\frac{n-4}{5}}$. I'm not getting from where does this $5$ in denominator came from..??
Let $u=a/b$, then
$$(a-b)^n=b^n(1-u)^n=b^n\sum_{k=0}^{n}(-u)^k \binom{n}{k}$$
The sum of 5th term and 6th term is:
$$b^n \left((-u)^4 \binom{n}{4}+(-u)^5 \binom{n}{5}\right)=0 \implies b^n u^4\left(\binom{n}{4}-u \binom{n}{5}\right)=0$$
Assume that $u \not = 0$ and $b\not = 0$ then we have
$$\frac{a}{b}=u=\frac{\binom{n}{4}}{\binom{n}{5}}=\frac{n!}{4!(n-4)!}\frac{5!(n-5)!}{n!}=\frac{5}{n-4}$$