left( {_{,4}^{47}} right) + sumlimits_{r = 1} | vedclass

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Question

$\left( {\begin{array}{*{20}{c}}
  {47} \ 
  4 
\end{array}} ight) + \sum\limits_{r = 1}^5 {\left( {\begin{array}{*{20}{c}}

Vedclass · Accepted Answer

$\left( {\begin{array}{*{20}{c}}{52}\4\end{array}} ight)$

Vedclass · Answer

$\left( {\begin{array}{*{20}{c}}
  {47} \ 
  4 
\end{array}} ight) + \sum\limits_{r = 1}^5 {\left( {\begin{array}{*{20}{c}}
  {52 - r} \ 
  3 
\end{array}} ight)} $

$ = \left( {\begin{array}{*{20}{c}}
  {51} \ 
  3 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {50} \ 
  3 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {49} \ 
  3 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {48} \ 
  3 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {47} \ 
  4 
\end{array}} ight)$

$\left( {\begin{array}{*{20}{c}}
  {51} \ 
  3 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {50} \ 
  3 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {49} \ 
  3 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {48} \ 
  3 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {48} \ 
  4 
\end{array}} ight)$

$ = \left( {\begin{array}{*{20}{c}}
  {51} \ 
  3 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {50} \ 
  3 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {49} \ 
  3 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {49} \ 
  4 
\end{array}} ight)$

$\left( {\begin{array}{*{20}{c}}
  {51} \ 
  3 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {50} \ 
  3 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {50} \ 
  4 
\end{array}} ight)$

$ = \left( {\begin{array}{*{20}{c}}
  {51} \ 
  3 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {51} \ 
  4 
\end{array}} ight) = \left( {\begin{array}{*{20}{c}}
  {52} \ 
  4 
\end{array}} ight)$

$\left( {_{\,4}^{47}} \right) + \sum\limits_{r = 1}^5 {\left( {_{\,\,\,\,3}^{52 - r}} \right)} = .........$

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