Correct Answer - Option 2 : 9
Concept:
General term: General term in the expansion of (x + y)n is given by \({{\rm{T}}_{\left( {{\rm{r\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}} \times {{\rm{x}}^{{\rm{n}} - {\rm{r}}}} \times {{\rm{y}}^{\rm{r}}}\)
Calculation:
We have to find 4th term in the expansion of \({\left( {\sqrt x + \;\frac{1}{x}} \right)^{n}}\)
We know that, \({{\rm{T}}_{\left( {{\rm{r\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}} \times {{\rm{x}}^{{\rm{n}} - {\rm{r}}}} \times {{\rm{y}}^{\rm{r}}}\)
\( \Rightarrow {T_4} = \;{{\rm{T}}_{\left( {3{\rm{\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{n}}{{\rm{C}}_3} \times {\left( {\sqrt {\rm{x}} } \right)^{n - 3}} \times {\left( {\frac{1}{x}} \right)^3}\)
= \({\;^{\rm{n}}}{{\rm{C}}_3} \times {\left( {\rm{x}} \right)^{\frac{{{\rm{n}} - 3}}{2}}} \times {\left( {\rm{x}} \right)^{ - 3}} \)
\(= {\;^{\rm{n}}}{{\rm{C}}_3} \times {\left( {\rm{x}} \right)^{\left( {\frac{{{\rm{n}} - 3}}{2} - 3} \right)}}\)
\( = {\;^{\rm{n}}}{{\rm{C}}_3} \times {\left( {\rm{x}} \right)^{\left( {\frac{{{\rm{n}} - 3 - 6}}{2}} \right)}} = {\;^{\rm{n}}}{{\rm{C}}_3} \times {\left( {\rm{x}} \right)^{\left( {\frac{{{\rm{n}} - 9}}{2}} \right)}}\)
4th term in the expansion of is independent of x,
Therefore, \(\rm \frac{n-9}{2} = 0\)
∴ n = 9
