Question

If the second term, in the expression ${\left[\sqrt[13]{a}+\frac{a}{\sqrt{a^{-1}}}\right]}^n$is $14a^{\frac{5}{2}}$, then the value of $\frac{n_{C_3}}{n_{C_2}}$ is:

• A. $4$
• B. $3$
• C. $12$
• D. $6$

Answer 1

The correct option is A

$4$

Explanation of correct option.

Step 1: Find the value of $n$

It is given that $T_{1+1}=14a^{\frac{5}{2}}$

Therefore,

$$\begin{gathered} n_{C_1}(13\sqrt{a}{)}^{n-1}\left(\frac{a}{\sqrt{a-1}}\right)=14a^{\frac{5}{2}} \\ \Rightarrow n\left(a^{\frac{n-1}{13}}{a}^{\frac{3}{2}}\right)=14a^{\frac{5}{2}} \\ \Rightarrow n\left(a^{\frac{n-1}{13}}{a}^{\frac{3}{2}}\right)=14a^{\frac{5}{2}} \\ \Rightarrow n\times a^{\frac{n-1}{13}+\frac{3}{2}}=14a^{\frac{5}{2}} \\ \Rightarrow n=14 \end{gathered}$$

Step 2. Find the value of $\frac{n_{C_3}}{n_{C_2}}$

$$\begin{gathered} \therefore \frac{n_{C_3}}{n_{C_2}} \\ =\frac{{}^{14}C_3}{{}^{14}C_2} \\ =\frac{{\displaystyle \frac{14!}{3!\left(14-3\right)!}}}{{\displaystyle \frac{14!}{2!\left(14-2!\right)}}} \\ =\frac{12}{3} \\ =4 \end{gathered}$$

Hence, Option (A) is correct.