A) \[\frac{n-4}{2}\]
B) \[\frac{n-4}{3}\]
C) \[\frac{n-4}{5}\]
D) \[\frac{n-4}{4}\]
Correct Answer: C
Solution :
Given expansion is \[{{(a-b)}^{n}}.\] \[\therefore \] \[{{T}_{5}}{{=}^{n}}{{C}_{4}}{{(a)}^{(n-4)}}{{(-b)}^{4}}\] and \[{{T}_{6}}{{=}^{n}}{{C}_{5}}{{(a)}^{(n-5)}}{{(-b)}^{5}}\] According to the given condition, \[{{T}_{5}}+{{T}_{6}}=0\] \[\therefore \] \[^{n}{{C}_{4}}{{(a)}^{n-4}}{{(-b)}^{4}}{{+}^{n}}{{C}_{5}}{{(a)}^{n-5}}{{(-b)}^{5}}=0\] \[\Rightarrow \] \[({{a}^{n-4}}){{(-b)}^{4}}\left[ ^{n}{{C}_{4}}{{+}^{n}}{{C}_{5}}\left( \frac{-b}{a} \right) \right]=0\] \[\Rightarrow \] \[\frac{a}{b}=\frac{^{n}{{C}_{5}}}{^{n}{{C}_{4}}}=\frac{\frac{n(n-1)\,(n-2)\,(n-3)\,(n-4)}{5\times 4\times 3\times 2\times 1}}{\frac{n(n-1)\,(n-2)\,(n-3)}{4\times 3\times 2\times 1}}\] \[=\frac{(n-4)}{5}\]
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