Question

If the second term of the expansion of ${{\left[ {{a}^{\dfrac{1}{13}}}+\dfrac{a}{\sqrt{{{a}^{-1}}}} \right]}^{n}}$ is $14{{a}^{\dfrac{5}{2}}}$ then the value of $\dfrac{{}^{n}{{C}_{3}}}{{}^{n}{{C}_{2}}}$ is

& \text{A}.\text{ 4} \\\ & \text{B}.\text{ 3} \\\ & \text{C}.\text{ 12} \\\ & \text{D}.\text{ 6} \\\ \end{aligned}$$

Answer 1

To solve this question, we will use 2 concepts. First the combination value ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and for the expansion of type ${{\left( a+b \right)}^{n}}$ the general term $T_{r+1}$ is given as ${}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$

Complete step-by-step solution:
Given that second term of expansion of
${{\left[ {{a}^{\dfrac{1}{13}}}+\dfrac{a}{\sqrt{{{a}^{-1}}}} \right]}^{n}}$ is $14{{a}^{\dfrac{5}{2}}}$
Consider ${{\left[ {{a}^{\dfrac{1}{13}}}+\dfrac{a}{\sqrt{{{a}^{-1}}}} \right]}^{n}}$
$\Rightarrow {{\left[ {{a}^{\dfrac{1}{13}}}+\dfrac{{{a}^{1}}}{{{a}^{\dfrac{-1}{2}}}} \right]}^{n}}={{\left[ {{a}^{\dfrac{1}{13}}}+{{a}^{\dfrac{3}{2}}} \right]}^{n}}$
Then, if the general term of an expansion $T_{r+1}$ is $T_{r+1}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$ where, ${{\left( a+b \right)}^{n}}$ expansion is taken. Then, general term for expansion of ${{\left[ {{a}^{\dfrac{1}{13}}}+{{a}^{\dfrac{3}{2}}} \right]}^{n}}$ is $Tr+1={}^{n}{{C}_{r}}{{a}^{\dfrac{n-r}{13}}}{{a}^{\dfrac{3r}{2}}}$
Given that ${{\text{T}}_{2}}\left( \text{second term} \right)=14{{a}^{\dfrac{5}{2}}}$
Then as $\text{r=1, }{{\text{T}}_{2}}={}^{n}{{C}_{1}}{{a}^{\dfrac{n-r}{13}}}{{a}^{\dfrac{3r}{2}}}=14{{a}^{\dfrac{5}{2}}}$

& \Rightarrow {}^{n}{{\text{C}}_{1}}{{\text{a}}^{\dfrac{n-1}{13}}}{{\text{a}}^{\dfrac{3}{2}}}=14{{\text{a}}^{\dfrac{5}{2}}} \\\ & \Rightarrow \text{n}{{\text{a}}^{\dfrac{n-1}{13}}}{{\text{a}}^{\dfrac{3}{2}}}=14{{\text{a}}^{\dfrac{5}{2}}} \\\ \end{aligned}$$ Now, taking powers of a together, we get, $$\begin{aligned} & \Rightarrow n{{a}^{\dfrac{39-2n-2}{26}}}=n{{a}^{\dfrac{37+2n}{26}}}=14{{a}^{\dfrac{5}{2}}} \\\ & \Rightarrow n{{a}^{\dfrac{37+2n}{26}}}=14{{a}^{\dfrac{5}{2}}} \\\ \end{aligned}$$ Taking 'a' on same side, we get: $$\begin{aligned} & 14=n {a}^{{\dfrac{37+2n}{26}}-{\dfrac{5}{2}}} \\\ & 14=n{{a}^{\dfrac{37+2n-65}{26}}} \\\ & 14=n{{a}^{\dfrac{2n-28}{26}}} \\\ \end{aligned}$$ Taking 2 common on powers of 'a' we get: $$14=n{{a}^{\dfrac{n-14}{13}}}$$ Comparing both sides of the equation we see that, there is no 'a' on the left-hand side of the above equation. $$\Rightarrow \text{ power of a=0 then }{{\text{a}}^{o}}=1$$ Hence, $$\begin{aligned} & \dfrac{n-14}{13}=0\Rightarrow n-14=0 \\\ & \Rightarrow n=14 \\\ \end{aligned}$$ So, value of n = 14. Now, finally we have to calculate $$\dfrac{{}^{n}{{C}_{3}}}{{}^{n}{{C}_{2}}}$$ $$\Rightarrow \dfrac{{}^{14}{{C}_{3}}}{{}^{14}{{C}_{2}}}$$ Now, we will use formula of $${}^{n}{{C}_{r}}$$ given as $${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$$ Using the formula stated above, we get $$\begin{aligned} & {}^{14}{{C}_{3}}=\dfrac{14!}{3!\left( 14-3 \right)!} \\\ & \Rightarrow {}^{14}{{C}_{3}}=\dfrac{14!}{3!\text{ }\times \text{ }11!} \\\ \end{aligned}$$ Opening on the factorial on to the right, we get $${}^{14}{{C}_{3}}=\dfrac{14\times 13\times 12\times 11!}{3\times 2\times 1\times 11!}$$ Cancelling 11! on both sides we have, $${}^{14}{{C}_{3}}=\dfrac{14\times 13\times 12}{3\times 2}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}$$ Similarly, we will calculate ${}^{14}{{C}_{2}}$ using formula stated above, Then, $$\begin{aligned} & {}^{14}{{C}_{2}}=\dfrac{14\times 13\times 12!}{\left( 14-2 \right)!\left( 2 \right)!} \\\ & \Rightarrow \dfrac{14\times 13\times 12!}{12!\times 2\times 1} \\\ \end{aligned}$$ Cancelling 12! we get, $${}^{14}{{C}_{2}}=\dfrac{14\times 13}{2\times 1}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}$$ Now from equation (i) and (ii) we get: $$\Rightarrow \dfrac{{}^{14}{{C}_{3}}}{{}^{14}{{C}_{2}}}=\dfrac{14\times 13\times 12}{3\times 2}\times \dfrac{2\times 1}{14\times 13}$$ Cancelling common terms we get: **$$\Rightarrow \dfrac{{}^{14}{{C}_{3}}}{{}^{14}{{C}_{2}}}=\dfrac{12}{3}=4$$ which is option A.** **Note:** The possibility of error can be at the point where we have to determine the value of n. Always in these types of questions, try to assemble the variable used at one side of the equation. Then, we can easily compare both sides of the equation to get an answer. Also for the second term of the expansion go for comparing terms of a and b from general term Tr+1 in the expansion of a and b.