$\sum\limits_{r = 0}^n {\,\frac{r}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  r 
\end{array}} \right)}}} \, = \,\frac{0}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  0 
\end{array}} \right)}}\, + \,\frac{1}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  1 
\end{array}} \right)}}\, + \,\frac{2}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  2 
\end{array}} \right)}}\, + \,…..\, + \,\frac{{n\, – \,1}}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  {n\, – \,1} 
\end{array}} \right)}}\, + \,\frac{n}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  n 
\end{array}} \right)}}$

$\sum\limits_{r = 0}^n {\,\frac{r}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  r 
\end{array}} \right)}}} \, = \,\frac{0}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  n 
\end{array}} \right)}}\, + \,\frac{1}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  {n\, – \,1} 
\end{array}} \right)}}\, + \,\frac{2}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  {n\, – \,2} 
\end{array}} \right)}}\, + \,……\, + \,\frac{{n\, – \,1}}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  1 
\end{array}} \right)}}\, + \,\frac{n}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  0 
\end{array}} \right)}}\,….\,(2)\,$

$\left[ {\because \,\left( {\begin{array}{*{20}{c}}
  n \\ 
  r 
\end{array}} \right)\, = \,\left( {\begin{array}{*{20}{c}}
  n \\ 
  {n\, – \,2} 
\end{array}} \right)} \right]$

હવે 

$\,{\text{(1)  +  (2) }}\, \Rightarrow \,2\,\sum\limits_{r\, = \,0}^n {\frac{r}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  r 
\end{array}} \right)}}\, = \,\frac{n}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  0 
\end{array}} \right)}}} \, + \,\frac{n}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  1 
\end{array}} \right)}}\, + \,……\, + \,\frac{n}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  n 
\end{array}} \right)}}$

$ = \,n\,\left[ {\frac{1}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  0 
\end{array}} \right)}}\, + \,\frac{1}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  1 
\end{array}} \right)}}\, + \,…..\, + \,\frac{1}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  n 
\end{array}} \right)}}} \right] = \,n\,\sum\limits_{r\, = \,0}^n {\frac{1}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  r 
\end{array}} \right)}}} \, = \,n{a_n}$

$\sum\limits_{r\, = 0}^n {\,\frac{r}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  r 
\end{array}} \right)}}} \, = \,\frac{n}{2}\,{a_n}$