If the second term of the expansion ;$\displaystyle \left [ a^{1/13}+\frac{a}{\sqrt{a^{-1}}} \right ]^{n}\: \: is\: \: 14a^{5/2}$, then the value of& ;$\displaystyle \frac{^{n}{C}_{3}}{^{n}{C}_{2}}$ is

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

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

$\Rightarrow { T }_{ 2 }={ _{ }^{ n }{ C } }_{ 1 }{ a }^{ \tfrac { n-1 }{ 13 } }{ a }^{ 1+\frac { 1 }{ 2 } }=14{ a }^{ \tfrac { 5 }{ 2 } }$

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

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

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

Since bases are same,we can equate the powers

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

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

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

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

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

$\cfrac { { _{ }^{ n }{ C } }_{ 3 } }{ { _{ }^{ n }{ C } }_{ 2 } } =\cfrac { 14-3+1 }{ 3 } =\cfrac { 12 }{ 3 } =4$