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$\left( {\begin{array}{*{20}{c}}n\\{r + 1}\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}n\\r\end{array}} \right) + \left( {\begin{array}{*{20}{c}}n\\{r - 1}\end{array}} \right) = .......$
$\left( {_{\,\,\,r}^{n + 2}} \right)$
$\left( {\,_{r + 1}^{n + 2}\,} \right)$
$\left( {_{\,\,\,r}^{n + 1}} \right)$
$\left( {\,_{r + 1}^{n + 1}\,} \right)$
Solution
$\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
n \\
{r – 1}
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
n \\
{r – 2}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{n + 2} \\
r
\end{array}} \right)\,\,$ થાય તે પ્રમાણે
$\left( {\begin{array}{*{20}{c}}
n \\
{r + 1}
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
n \\
{r – 1}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{n + 2} \\
{r + 1}
\end{array}} \right)\,$ મળે
