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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.

The answer is: (n-4)/5

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  • $\begingroup$ Note $$b^{-n}(a-b)^n=\left({a\over b}-1\right)^n$$ $\endgroup$ Sep 20 '14 at 20:36
  • $\begingroup$ You forgot the signs in the terms. $\endgroup$ Sep 20 '14 at 20:39
  • $\begingroup$ You also need to assume $a,b\neq 0$... $\endgroup$ Sep 20 '14 at 20:40
  • $\begingroup$ @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? $\endgroup$
    – Swetank
    Sep 20 '14 at 20:47
  • $\begingroup$ @ThomasAndrews a,b ≠ 0? Didn't get that. $\endgroup$
    – Swetank
    Sep 20 '14 at 20:52
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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.

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  • $\begingroup$ And so I have come to, a/b = nC5/nC4. How to solve further? $\endgroup$
    – Swetank
    Sep 20 '14 at 21:57
  • $\begingroup$ You arrive to $\frac{a}{b}=\frac{\binom{n}5} {\binom{n}4}$. Now express the binomial numbers using factorials: remind that $\binom{n}k=\frac {n!}{k!(n-k)!}$. $\endgroup$
    – Anatoly
    Sep 20 '14 at 22:04
  • $\begingroup$ a/b = (n-4)!/5(n-5)! .. I haven't learnt how to solve them further. Can you guide me through? $\endgroup$
    – Swetank
    Sep 20 '14 at 22:10
  • $\begingroup$ $$\frac{\binom {n}5}{ \binom {n}4}= \frac {n!}{5!(n-5)!} \frac {4!(n-4)!}{n!}=\frac {4!}{5!} \frac {(n-4)!}{(n-5)!}=\frac {n-4}{5}$$ $\endgroup$
    – Anatoly
    Sep 20 '14 at 22:17
  • $\begingroup$ I get 4!/5! = 1/5. But How (n-4)!/(n-5)! = n-4? I seriously had no experience on that. This is now the only part which is confusing me a lot. Please help me with that. $\endgroup$
    – Swetank
    Sep 20 '14 at 22:21

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