Question

If the second term in the expansion of ${(\sqrt[13]{13a}+a^{\frac{3}{2}})}^n$ is $14a^{\frac{5}{2}}$, then the value of $\frac{{}^n{C}_3}{{}^n{C}_3}$ is:

• A. $3$
• B. $1$
• C. $6$
• D. $12$

Answer 1

The correct option is B $1$

${(a^{\frac{1}{13}}+a^{\frac{3}{2}})}^n$

$T_{r+1}{=}^n{C}_r{a}^{\frac{n-r}{13}+\frac{3r}{2}}$

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

According to question

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

on comparing

${}^n{C}_1=14$

n=14

$\frac{n-1}{13}+\frac{3}{2}=\frac{5}{2}$

n=14

but $\frac{{}^n{C}_3}{{}^n{C}_3}=1$