In the binomial expansion of (a-b)^n, n5, the sum of 5th and 6th terms is zero, then a/b e…

Question

In the binomial expansion of $(a-b)^n$, $n\ge5$, the sum of 5th and 6th terms is zero, then $a/b$ equals:

Accepted Answer

$T_5={}^nC_4a^{n-4}b^4$, $T_6=-{}^nC_5a^{n-5}b^5$. Given $T_5+T_6=0$:

$${}^nC_4a^{n-4}b^4={}^nC_5a^{n-5}b^5\quad\Rightarrow\quad\frac{1}{n-4}=\frac{b}{5a}\quad\Rightarrow\quad\frac ab=\frac{n-4}{5}$$

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