Question:
In the binomial expansion of $ {{(a-b)}^{n}},\,\,n\,\,\ge \,5, $ the sum of $5^{th}$ and $6^{th}$ term is zero. Then, $ \frac{a}{b} $ is equal to
In the binomial expansion of $ {{(a-b)}^{n}},\,\,n\,\,\ge \,5, $ the sum of $5^{th}$ and $6^{th}$ term is zero. Then, $ \frac{a}{b} $ is equal to
Updated On: Jun 23, 2024
- $ \frac{n-4}{2} $
- $ \frac{n-4}{3} $
- $ \frac{n-4}{5} $
- $ \frac{n-4}{4} $
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The Correct Option is C
Solution and Explanation
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} $
$ \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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Concepts Used:
Binomial Theorem
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
Properties of Binomial Theorem
- The number of coefficients in the binomial expansion of (x + y)n is equal to (n + 1).
- There are (n+1) terms in the expansion of (x+y)n.
- The first and the last terms are xn and yn respectively.
- From the beginning of the expansion, the powers of x, decrease from n up to 0, and the powers of a, increase from 0 up to n.
- The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.