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જો $\left( {\begin{array}{*{20}{c}}
{n\, - \,1} \\
r
\end{array}} \right)\,\, = \,\,\left( {\,{k^2}\, - \,3\,} \right)\,\,\left( {\begin{array}{*{20}{c}}
n \\
{r\, + \,1}
\end{array}} \right)\,$ તો $k\, \in \,\,..........$
$[ - 2, - \sqrt 3 ]\,\, \cup \,\,[\sqrt 3 ,2]$
$( - 2, - \sqrt 3 )\,\, \cup \,\,(\sqrt 3 ,2)$
$( - 2, - \sqrt 3 ]\,\, \cup \,\,[\sqrt 3 ,2)$
$[ - 2,\sqrt 3 )\,\, \cup \,\,(\sqrt 3 ,2]$
Solution
$\left( {\begin{array}{*{20}{c}}
{n\, – \,1} \\
r
\end{array}} \right)\, = \,\left( {\,{k^2}\, – \,3\,} \right)\,\left( {\begin{array}{*{20}{c}}
n \\
{r\, + \,1}
\end{array}} \right)$
$\frac{{\left( {\begin{array}{*{20}{c}}
{n\, – \,1} \\
r
\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}
n \\
{r\, + \,1}
\end{array}} \right)}}\, = \,{k^2}\, – \,3\,$
$\frac{{(n\, – \,1)!}}{{(n\, – \,1\, – \,r)\,!\,r\,!}}\, \times \,\frac{{(n\, – \,r\, – \,1)\,!\,(r\, + \,1)!}}{{n!}}\, = \,{k^2}\, – \,3\,\,$
હવે $0 \leq r \leq n – 1 $ $ 1 \leq r + 1 \leq n $
$\frac{1}{n}\, \leqslant \,\frac{{r\, + \,1}}{n}\, \leqslant \,1\,\,\,$
$\,0\, < \,\frac{{r\, + \,1}}{n}\, \leqslant \,1$
$3 < {k^2} \leqslant 4\,$
$k \in [ – 2, – \sqrt 3 ) \cup (\sqrt 3 ,2]$