Question

If ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}}$ are coefficients of ${{2}^{nd}},{{3}^{rd}},{{4}^{th}}$ and ${{5}^{th}}$ terms in the expansion of ${{\left( 1+x \right)}^{n}}$ respectively, then $\dfrac{{{a}_{1}}}{{{a}_{1}}+{{a}_{2}}}+\dfrac{{{a}_{3}}}{{{a}_{3}}+{{a}_{4}}}$
A. $\dfrac{{{a}_{2}}}{{{a}_{2}}+{{a}_{3}}}$
B. $\dfrac{2{{a}_{2}}}{{{a}_{2}}+{{a}_{3}}}$
C. $\dfrac{3{{a}_{2}}}{{{a}_{2}}+{{a}_{3}}}$
D. $\dfrac{4{{a}_{3}}}{{{a}_{2}}+{{a}_{3}}}$

Answer 1

We will first find out the coefficients ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}}$in the expansion of ${{\left( 1+x \right)}^{n}}$by the formula ${{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{x}^{r}}$where ${{T}_{r+1}}$ is the ${{\left( r+1 \right)}^{th}}$ term in the expansion. After finding their values, we will put them in our given expression and obtain its value. Then we will expand the and find the values of the expressions given in the options and hence we will get our answer.

Complete step-by-step answer:
Now, we know that the ${{\left( r+1 \right)}^{th}}$ term in the expansion of ${{\left( 1+x \right)}^{n}}$ is given as:
${{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{x}^{r}}$
Thus, the coefficient of the ${{\left( r+1 \right)}^{th}}$term in the expansion of ${{\left( 1+x \right)}^{n}}$is given as:
${{a}_{r+1}}{{=}^{n}}{{C}_{r}}$
Thus, the coefficients of the ${{2}^{nd}},{{3}^{rd}},{{4}^{th}}$ and ${{5}^{th}}$ term in the expansion of ${{\left( 1+x \right)}^{n}}$ are given as:

& {{a}_{1}}{{=}^{n}}{{C}_{1}} \\\ & {{a}_{2}}{{=}^{n}}{{C}_{2}} \\\ & {{a}_{3}}{{=}^{n}}{{C}_{3}} \\\ & {{a}_{4}}{{=}^{n}}{{C}_{4}} \\\ \end{aligned}$$ Now, we have to find the value of the expression $\dfrac{{{a}_{1}}}{{{a}_{1}}+{{a}_{2}}}+\dfrac{{{a}_{3}}}{{{a}_{3}}+{{a}_{4}}}$. By putting in the obtained values in the given expression, we will get our required value. Putting in the values we get: $\begin{aligned} & \dfrac{{{a}_{1}}}{{{a}_{1}}+{{a}_{2}}}+\dfrac{{{a}_{3}}}{{{a}_{3}}+{{a}_{4}}} \\\ & \Rightarrow \dfrac{^{n}{{C}_{1}}}{^{n}{{C}_{1}}{{+}^{n}}{{C}_{2}}}+\dfrac{^{n}{{C}_{3}}}{^{n}{{C}_{3}}{{+}^{n}}{{C}_{4}}} \\\ \end{aligned}$ Now, we know the property that: $^{n}{{C}_{r}}{{+}^{n}}{{C}_{r+1}}{{=}^{n+1}}{{C}_{r+1}}$ This property is applicable in the denominators. Applying this property, we get: $\begin{aligned} & \dfrac{^{n}{{C}_{1}}}{^{n}{{C}_{1}}{{+}^{n}}{{C}_{2}}}+\dfrac{^{n}{{C}_{3}}}{^{n}{{C}_{3}}{{+}^{n}}{{C}_{4}}} \\\ & \Rightarrow \dfrac{^{n}{{C}_{1}}}{^{n+1}{{C}_{2}}}+\dfrac{^{n}{{C}_{3}}}{^{n+1}{{C}_{4}}} \\\ \end{aligned}$ Now, we know that the expansion of $^{n}{{C}_{r}}$ is given as: $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ Thus, we will now open the expansions in our expression. Opening the values in the expression we get: $$\begin{aligned} & \dfrac{^{n}{{C}_{1}}}{^{n+1}{{C}_{2}}}+\dfrac{^{n}{{C}_{3}}}{^{n+1}{{C}_{4}}} \\\ & \Rightarrow \dfrac{\dfrac{n!}{1!\left( n-1 \right)!}}{\dfrac{\left( n+1 \right)!}{2!\left( n-1 \right)!}}+\dfrac{\dfrac{n!}{3!\left( n-3 \right)!}}{\dfrac{\left( n+1 \right)!}{4!\left( n-3 \right)!}} \\\ & \Rightarrow \dfrac{n!}{\dfrac{\left( n+1 \right)!}{2}}+\dfrac{n!}{\dfrac{\left( n+1 \right)!}{4}} \\\ & \Rightarrow \dfrac{2n!}{\left( n+1 \right)!}+\dfrac{4n!}{\left( n+1 \right)!} \\\ & \Rightarrow \dfrac{6n!}{\left( n+1 \right)!} \\\ & \Rightarrow \dfrac{6n!}{\left( n+1 \right)n!} \\\ & \Rightarrow \dfrac{6}{n+1} \\\ \end{aligned}$$ Thus, the value of the required expression is $\dfrac{6}{n+1}$ . Now, most of our given options are integral multiples of $\dfrac{{{a}_{2}}}{{{a}_{2}}+{{a}_{3}}}$ Thus, we will check the value of $\dfrac{{{a}_{2}}}{{{a}_{2}}+{{a}_{3}}}$and then see which of its integral multiple will give us our required value. Now, putting in the values in the expression $\dfrac{{{a}_{2}}}{{{a}_{2}}+{{a}_{3}}}$, we get: $\begin{aligned} & \dfrac{{{a}_{2}}}{{{a}_{2}}+{{a}_{3}}} \\\ & \Rightarrow \dfrac{^{n}{{C}_{2}}}{^{n}{{C}_{2}}{{+}^{n}}{{C}_{3}}} \\\ \end{aligned}$ Using the property $^{n}{{C}_{r}}{{+}^{n}}{{C}_{r+1}}{{=}^{n+1}}{{C}_{r+1}}$ in the denominator, we get: $\begin{aligned} & \dfrac{^{n}{{C}_{2}}}{^{n}{{C}_{2}}{{+}^{n}}{{C}_{3}}} \\\ & \Rightarrow \dfrac{^{n}{{C}_{2}}}{^{n+1}{{C}_{3}}} \\\ \end{aligned}$ Now, opening the expressions we get: $\begin{aligned} & \dfrac{^{n}{{C}_{2}}}{^{n+1}{{C}_{3}}} \\\ & \Rightarrow \dfrac{\dfrac{n!}{2!\left( n-2 \right)!}}{\dfrac{\left( n+1 \right)!}{3!\left( n-2 \right)!}} \\\ & \Rightarrow \dfrac{n!}{\dfrac{\left( n+1 \right)!}{3}} \\\ & \Rightarrow \dfrac{3n!}{\left( n+1 \right)!} \\\ & \Rightarrow \dfrac{3n!}{\left( n+1 \right)n!} \\\ & \Rightarrow \dfrac{3}{n+1} \\\ \end{aligned}$ Now, our required value is $\dfrac{6}{n+1}$ and the value of $\dfrac{{{a}_{2}}}{{{a}_{2}}+{{a}_{3}}}$ is $\dfrac{3}{n+1}$. Thus, our required expression is twice of $\dfrac{{{a}_{2}}}{{{a}_{2}}+{{a}_{3}}}$. Hence, our answer is the second integral multiple of $\dfrac{{{a}_{2}}}{{{a}_{2}}+{{a}_{3}}}$, i.e. $\dfrac{2{{a}_{2}}}{{{a}_{2}}+{{a}_{3}}}$ **So, the correct answer is “Option B”.** **Note:** After obtaining the value of our required expression, we checked for the value of $\dfrac{{{a}_{2}}}{{{a}_{2}}+{{a}_{3}}}$. We also could have checked the options one by one but it would be more time taking and also we would have done the same calculation again and again. The expression $\dfrac{{{a}_{2}}}{{{a}_{2}}+{{a}_{3}}}$is there in most of the options, so we checked it instead to save time.