# Binomial Expansion of $(a-b)^n$ Find value of $\frac ab$.

Q: In the binomial expansion of $(a-b)^n, n\geq 5$, the sum of the $5$th and $6$th term is $0$. Find the value of $\frac{a}{b}$.

I had found 5th and 6th term. Which is:

• 5th term: $\binom{n}4\cdot a^{n-4} \cdot b^4$

• 6th term: $\binom{n}5 \cdot a^{n-5} \cdot -b^5$

I don't know how to proceed further. Please don't post answers directly, help me with some steps first.

• Note $$b^{-n}(a-b)^n=\left({a\over b}-1\right)^n$$ Sep 20 '14 at 20:36
• You forgot the signs in the terms. Sep 20 '14 at 20:39
• You also need to assume $a,b\neq 0$... Sep 20 '14 at 20:40
• @ThomasAndrews Thanks, I have corrected the signs. so, now term 5th - 6th = 0. How to cancel out other terms, so that to get a/b? Sep 20 '14 at 20:47
• @ThomasAndrews a,b ≠ 0? Didn't get that. Sep 20 '14 at 20:52

Hint: equalize the two expressions that you have already found for the $5^{th}$ and $6^{th}$ terms, and solve the resulting equation for $\frac{a}{b}$. In doing this, write the binomial numbers using factorials.
• You arrive to $\frac{a}{b}=\frac{\binom{n}5} {\binom{n}4}$. Now express the binomial numbers using factorials: remind that $\binom{n}k=\frac {n!}{k!(n-k)!}$. Sep 20 '14 at 22:04
• $$\frac{\binom {n}5}{ \binom {n}4}= \frac {n!}{5!(n-5)!} \frac {4!(n-4)!}{n!}=\frac {4!}{5!} \frac {(n-4)!}{(n-5)!}=\frac {n-4}{5}$$ Sep 20 '14 at 22:17