Question

If the second term of the expansion ${[a^{1/13}+\frac{a}{\sqrt{a^{-1}}}]}^{n}is14a^{5/2}$, then the value of $\frac{{}^n{C}_3}{{}^n{C}_2}$ is

• A. $4$
• B. $3$
• C. $12$
• D. $6$

Answer 1

Answer: (A)

$4$

Using the formula $T_{r+1=}{{}^{n}C}_r{x}^{n-r}{a}^r$

in the above question,we have $T_2={{}^{n}C}_1{\left(a^{\frac{1}{13}}\right)}^{n-1}\left(a\sqrt{a}\right)=14a^{\frac{5}{2}}$

$\Rightarrow T_2={{}^{n}C}_1{a}^{\frac{n-1}{13}}{a}^{1+\frac{1}{2}}=14a^{\frac{5}{2}}$

$\Rightarrow {{}^{n}C}_1{a}^{\frac{n-1}{13}+\frac{3}{2}}=14a^{\frac{5}{2}}$

Comparing L.H.S and R.H.S,

$\Rightarrow a^{\frac{2n+37}{26}}=a^{\frac{5}{2}}$

Since bases are same,we can equate the powers

Thus, $\frac{2n+37}{26}=\frac{5}{2}$

On solving , $2(2n+37)=5\times 26\Rightarrow 2n+37=65$

$\therefore 2n=65-37=28$

Hence, $n=\frac{28}{2}=14$

Using the formula, $\frac{{{}^{n}C}_r}{{{}^{n}C}_{r-1}}=\frac{n-r+1}{r}$ where $r=3$ and $n=14$

$\frac{{{}^{n}C}_3}{{{}^{n}C}_2}=\frac{14-3+1}{3}=\frac{12}{3}=4$