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In the binomial expression of ${{\left( a-b \right)}^{n}},n>5$, the sum of ${{5}^{th}}$ and ${{6}^{th}}$ terms is zero, then $\dfrac{a}{b}$ equals-
A. $\dfrac{5}{n-5}$
B. $\dfrac{6}{n-5}$
C. $\dfrac{n-5}{6}$
D. $\dfrac{n-4}{5}$

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Last updated date: 25th Jul 2024
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Answer
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Hint: For finding the required ratio, we need to use the formula of binomial expression for ${{\left( a-b \right)}^{n}}$. After expansion of the binomial formula, we will choose ${{5}^{th}}$ and ${{6}^{th}}$ terms. Then, we will use the given condition in the question and it will provide us an equation. So, we will simplify that equation to get the required ratio.

Complete step-by-step solution:
Since, the binomial expression ${{\left( a-b \right)}^{n}}$ is given in the question. So, the expansion of binomial expression for up to 6 terms:
\[\Rightarrow {{\left( a-b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{(-b)}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{(-b)}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{(-b)}^{2}}+{}^{n}{{C}_{3}}{{a}^{n-3}}{{(-b)}^{3}}+{}^{n}{{C}_{4}}{{a}^{n-4}}{{(-b)}^{4}}+{}^{n}{{C}_{5}}{{a}^{n-5}}{{(-b)}^{5}}\]
Now, we will take ${{5}^{th}}$ and ${{6}^{th}}$ terms as:
$\Rightarrow {{5}^{th}}\text{term}={}^{n}{{C}_{4}}{{a}^{n-4}}{{(-b)}^{4}}$
$\Rightarrow {{6}^{th}}\text{term}={}^{n}{{C}_{5}}{{a}^{n-5}}{{(-b)}^{5}}$
Here, we will use the given condition in the question that the sum of ${{5}^{th}}$ and ${{6}^{th}}$ terms is zero as:
$\Rightarrow {}^{n}{{C}_{4}}{{a}^{n-4}}{{(-b)}^{4}}+{}^{n}{{C}_{5}}{{a}^{n-5}}{{(-b)}^{5}}=0$
Now, we will expand the every term of above step to make calculation easy as:
$\Rightarrow \dfrac{n!}{4!\centerdot \left( n-4 \right)!}\centerdot \dfrac{{{a}^{n}}}{{{a}^{4}}}\centerdot {{b}^{4}}-\dfrac{n!}{5!\centerdot \left( n-5 \right)!}\centerdot \dfrac{{{a}^{n}}}{{{a}^{5}}}\centerdot {{b}^{5}}=0$
Here, we can write the above term as:
$\Rightarrow \dfrac{n!}{4!\centerdot \left( n-4 \right)!}\centerdot \dfrac{{{a}^{n}}}{{{a}^{4}}}\centerdot {{b}^{4}}=\dfrac{n!}{5!\centerdot \left( n-5 \right)!}\centerdot \dfrac{{{a}^{n}}}{{{a}^{5}}}\centerdot {{b}^{5}}$
In the above step, we can cancel out the equal like terms as:
$\Rightarrow \dfrac{1}{4!\centerdot \left( n-4 \right)!}\centerdot \dfrac{1}{{{a}^{4}}}\centerdot {{b}^{4}}=\dfrac{1}{5!\centerdot \left( n-5 \right)!}\centerdot \dfrac{1}{{{a}^{5}}}\centerdot {{b}^{5}}$
Again we will write the above step in the simple form to make calculation easy as:
$\Rightarrow \dfrac{1}{4!\centerdot \left( n-4 \right)\centerdot \left( n-5 \right)!}\centerdot \dfrac{1}{{{a}^{4}}}\centerdot {{b}^{4}}=\dfrac{1}{5\centerdot 4!\centerdot \left( n-5 \right)!}\centerdot \dfrac{1}{a\centerdot {{a}^{4}}}\centerdot b\centerdot {{b}^{4}}$
Here, we will again cancel out the equal terms as:
\[\Rightarrow \dfrac{1}{\left( n-4 \right)}=\dfrac{1}{5}\centerdot \dfrac{b}{a}\]
We can write the above term as:
\[\Rightarrow \dfrac{5}{\left( n-4 \right)}=\dfrac{b}{a}\]
The above step can be written as:
\[\Rightarrow \dfrac{a}{b}=\dfrac{\left( n-4 \right)}{5}\]
Hence, In the binomial expression of ${{\left( a-b \right)}^{n}},n>5$, the sum of ${{5}^{th}}$ and ${{6}^{th}}$ terms is zero, then $\dfrac{a}{b}$ equals to \[\dfrac{\left( n-4 \right)}{5}\].

Note: In a binomial experiment, there are \[\text{n}\] independent trials and each trial has only two possible outcomes. One is success and another one is failure. If the probability of success is the same for each trial, this probability is denoted by \[\text{p}\] and the probability of failure is \[\left( \text{1}-\text{p} \right)\].
For a binomial experiment, the probability of exactly $r$ successes in \[\text{n}\] trials is \[\text{P}\left( \text{k successes} \right)\text{ }=\text{ }{}^{n}{{\text{C}}_{r}}\text{ }{{\text{p}}^{r}}{{\left( \text{1 }-\text{ p} \right)}^{n-r}}\]