Evaluate:$frac{1}{2^{101}}sum_{k=1}^{51} sum_{r=0}^{k-1}binom{51}{k}binom{50}{r}$












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$begingroup$


Evaluate:$$frac{1}{2^{101}}sum_{k=1}^{51} sum_{r=0}^{k-1}binom{51}{k}binom{50}{r}$$
My Attempt:



I did try writing the series $(1+x)^{50}$and $(1+x)^{50}$ separately,then multiplied but could not determine the power of $x$ whose coefficient should be found.Can Vandermonde's Identity be put to use here.






binomial-coefficients binomial-theorem binomial-distribution







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    3












    $begingroup$


    Evaluate:$$frac{1}{2^{101}}sum_{k=1}^{51} sum_{r=0}^{k-1}binom{51}{k}binom{50}{r}$$
    My Attempt:



    I did try writing the series $(1+x)^{50}$and $(1+x)^{50}$ separately,then multiplied but could not determine the power of $x$ whose coefficient should be found.Can Vandermonde's Identity be put to use here.






    binomial-coefficients binomial-theorem binomial-distribution







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      0



      $begingroup$


      Evaluate:$$frac{1}{2^{101}}sum_{k=1}^{51} sum_{r=0}^{k-1}binom{51}{k}binom{50}{r}$$
      My Attempt:



      I did try writing the series $(1+x)^{50}$and $(1+x)^{50}$ separately,then multiplied but could not determine the power of $x$ whose coefficient should be found.Can Vandermonde's Identity be put to use here.






      binomial-coefficients binomial-theorem binomial-distribution







      share|cite|improve this question









      $endgroup$




      Evaluate:$$frac{1}{2^{101}}sum_{k=1}^{51} sum_{r=0}^{k-1}binom{51}{k}binom{50}{r}$$
      My Attempt:



      I did try writing the series $(1+x)^{50}$and $(1+x)^{50}$ separately,then multiplied but could not determine the power of $x$ whose coefficient should be found.Can Vandermonde's Identity be put to use here.






      binomial-coefficients binomial-theorem binomial-distribution




      binomial-coefficients binomial-theorem binomial-distribution






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 16 '18 at 5:39









      MaverickMaverick

      2,101621




      2,101621






















          2 Answers
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          $begingroup$

          Using the identity $$
          binom{51}{k}=binom{50}{k}+binom{50}{k-1},
          $$           we get
          $$
          sum_{k=1}^{51} sum_{r=0}^{k-1}binom{51}{k}binom{50}{r}=sum_{k=1}^{51} sum_{r=0}^{k-1}binom{50}{k}binom{50}{r}+sum_{k=1}^{51} sum_{r=0}^{k-1}binom{50}{k-1}binom{50}{r}.$$           Observe that the former is equal to
          $$
          sum_{k=0}^{50} sum_{r=0}^{k-1}binom{50}{k}binom{50}{r}= sum_{ 0leq r<kleq 50}binom{50}{k}binom{50}{r} = sum_{ 0leq k<rleq 50}binom{50}{k}binom{50}{r}=sum_{k=0}^{50} sum_{r=k+1}^{50}binom{50}{k}binom{50}{r}.
          $$           Also note that the latter is equal to
          $$begin{eqnarray}
          sum_{k=1}^{51} sum_{r=0}^{k-1}binom{50}{k-1}binom{50}{r}=sum_{k=0}^{50}sum_{r=0}^{k}binom{50}{k}binom{50}{r}.
          end{eqnarray}$$           Gathering them together yields
          $$
          sum_{k=0}^{50} sum_{r=k+1}^{50}binom{50}{k}binom{50}{r}+sum_{k=0}^{50}sum_{r=0}^{k}binom{50}{k}binom{50}{r}=sum_{k=0}^{50} sum_{r=0}^{50}binom{50}{k}binom{50}{r} = 2^{50}cdot2^{50} = 2^{100}.
          $$           Hence the answer is $frac{1}{2}$.






          share|cite|improve this answer









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            2












            $begingroup$

            Here's an alternative solution.



            Observe
            begin{align}
            S=& sum^{51}_{k=1}sum^{k-1}_{r=0}binom{51}{k}binom{50}{r} = sum^{51}_{k=1}sum^{k-1}_{r=0}binom{51}{51-k}binom{50}{50-r}\
            =& sum^{50}_{k'=0} sum^{50}_{r'=k'}binom{51}{k'}binom{50}{r'} = sum^{51}_{k'=0} sum^{50}_{r'=k'}binom{51}{k'}binom{50}{r'}.
            end{align}
            Hence it follows
            begin{align}
            2S= sum^{51}_{k=0}sum_{r=0}^{50}binom{51}{k}binom{50}{r}= sum^{51}_{k=0}binom{51}{k}sum_{r=0}^{50}binom{50}{r}=2^{51}cdot 2^{50}
            end{align}
            which means
            begin{align}
            S = 2^{100}.
            end{align}
            Hence you arrive at $S/2^{101} = 1/2$.






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              2 Answers
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              3












              $begingroup$

              Using the identity $$
              binom{51}{k}=binom{50}{k}+binom{50}{k-1},
              $$               we get
              $$
              sum_{k=1}^{51} sum_{r=0}^{k-1}binom{51}{k}binom{50}{r}=sum_{k=1}^{51} sum_{r=0}^{k-1}binom{50}{k}binom{50}{r}+sum_{k=1}^{51} sum_{r=0}^{k-1}binom{50}{k-1}binom{50}{r}.$$               Observe that the former is equal to
              $$
              sum_{k=0}^{50} sum_{r=0}^{k-1}binom{50}{k}binom{50}{r}= sum_{ 0leq r<kleq 50}binom{50}{k}binom{50}{r} = sum_{ 0leq k<rleq 50}binom{50}{k}binom{50}{r}=sum_{k=0}^{50} sum_{r=k+1}^{50}binom{50}{k}binom{50}{r}.
              $$               Also note that the latter is equal to
              $$begin{eqnarray}
              sum_{k=1}^{51} sum_{r=0}^{k-1}binom{50}{k-1}binom{50}{r}=sum_{k=0}^{50}sum_{r=0}^{k}binom{50}{k}binom{50}{r}.
              end{eqnarray}$$               Gathering them together yields
              $$
              sum_{k=0}^{50} sum_{r=k+1}^{50}binom{50}{k}binom{50}{r}+sum_{k=0}^{50}sum_{r=0}^{k}binom{50}{k}binom{50}{r}=sum_{k=0}^{50} sum_{r=0}^{50}binom{50}{k}binom{50}{r} = 2^{50}cdot2^{50} = 2^{100}.
              $$               Hence the answer is $frac{1}{2}$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Using the identity $$
                binom{51}{k}=binom{50}{k}+binom{50}{k-1},
                $$                 we get
                $$
                sum_{k=1}^{51} sum_{r=0}^{k-1}binom{51}{k}binom{50}{r}=sum_{k=1}^{51} sum_{r=0}^{k-1}binom{50}{k}binom{50}{r}+sum_{k=1}^{51} sum_{r=0}^{k-1}binom{50}{k-1}binom{50}{r}.$$                 Observe that the former is equal to
                $$
                sum_{k=0}^{50} sum_{r=0}^{k-1}binom{50}{k}binom{50}{r}= sum_{ 0leq r<kleq 50}binom{50}{k}binom{50}{r} = sum_{ 0leq k<rleq 50}binom{50}{k}binom{50}{r}=sum_{k=0}^{50} sum_{r=k+1}^{50}binom{50}{k}binom{50}{r}.
                $$                 Also note that the latter is equal to
                $$begin{eqnarray}
                sum_{k=1}^{51} sum_{r=0}^{k-1}binom{50}{k-1}binom{50}{r}=sum_{k=0}^{50}sum_{r=0}^{k}binom{50}{k}binom{50}{r}.
                end{eqnarray}$$                 Gathering them together yields
                $$
                sum_{k=0}^{50} sum_{r=k+1}^{50}binom{50}{k}binom{50}{r}+sum_{k=0}^{50}sum_{r=0}^{k}binom{50}{k}binom{50}{r}=sum_{k=0}^{50} sum_{r=0}^{50}binom{50}{k}binom{50}{r} = 2^{50}cdot2^{50} = 2^{100}.
                $$                 Hence the answer is $frac{1}{2}$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Using the identity $$
                  binom{51}{k}=binom{50}{k}+binom{50}{k-1},
                  $$                   we get
                  $$
                  sum_{k=1}^{51} sum_{r=0}^{k-1}binom{51}{k}binom{50}{r}=sum_{k=1}^{51} sum_{r=0}^{k-1}binom{50}{k}binom{50}{r}+sum_{k=1}^{51} sum_{r=0}^{k-1}binom{50}{k-1}binom{50}{r}.$$                   Observe that the former is equal to
                  $$
                  sum_{k=0}^{50} sum_{r=0}^{k-1}binom{50}{k}binom{50}{r}= sum_{ 0leq r<kleq 50}binom{50}{k}binom{50}{r} = sum_{ 0leq k<rleq 50}binom{50}{k}binom{50}{r}=sum_{k=0}^{50} sum_{r=k+1}^{50}binom{50}{k}binom{50}{r}.
                  $$                   Also note that the latter is equal to
                  $$begin{eqnarray}
                  sum_{k=1}^{51} sum_{r=0}^{k-1}binom{50}{k-1}binom{50}{r}=sum_{k=0}^{50}sum_{r=0}^{k}binom{50}{k}binom{50}{r}.
                  end{eqnarray}$$                   Gathering them together yields
                  $$
                  sum_{k=0}^{50} sum_{r=k+1}^{50}binom{50}{k}binom{50}{r}+sum_{k=0}^{50}sum_{r=0}^{k}binom{50}{k}binom{50}{r}=sum_{k=0}^{50} sum_{r=0}^{50}binom{50}{k}binom{50}{r} = 2^{50}cdot2^{50} = 2^{100}.
                  $$                   Hence the answer is $frac{1}{2}$.






                  share|cite|improve this answer









                  $endgroup$



                  Using the identity $$
                  binom{51}{k}=binom{50}{k}+binom{50}{k-1},
                  $$                   we get
                  $$
                  sum_{k=1}^{51} sum_{r=0}^{k-1}binom{51}{k}binom{50}{r}=sum_{k=1}^{51} sum_{r=0}^{k-1}binom{50}{k}binom{50}{r}+sum_{k=1}^{51} sum_{r=0}^{k-1}binom{50}{k-1}binom{50}{r}.$$                   Observe that the former is equal to
                  $$
                  sum_{k=0}^{50} sum_{r=0}^{k-1}binom{50}{k}binom{50}{r}= sum_{ 0leq r<kleq 50}binom{50}{k}binom{50}{r} = sum_{ 0leq k<rleq 50}binom{50}{k}binom{50}{r}=sum_{k=0}^{50} sum_{r=k+1}^{50}binom{50}{k}binom{50}{r}.
                  $$                   Also note that the latter is equal to
                  $$begin{eqnarray}
                  sum_{k=1}^{51} sum_{r=0}^{k-1}binom{50}{k-1}binom{50}{r}=sum_{k=0}^{50}sum_{r=0}^{k}binom{50}{k}binom{50}{r}.
                  end{eqnarray}$$                   Gathering them together yields
                  $$
                  sum_{k=0}^{50} sum_{r=k+1}^{50}binom{50}{k}binom{50}{r}+sum_{k=0}^{50}sum_{r=0}^{k}binom{50}{k}binom{50}{r}=sum_{k=0}^{50} sum_{r=0}^{50}binom{50}{k}binom{50}{r} = 2^{50}cdot2^{50} = 2^{100}.
                  $$                   Hence the answer is $frac{1}{2}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 16 '18 at 6:41









                  SongSong

                  16.8k21145




                  16.8k21145























                      2












                      $begingroup$

                      Here's an alternative solution.



                      Observe
                      begin{align}
                      S=& sum^{51}_{k=1}sum^{k-1}_{r=0}binom{51}{k}binom{50}{r} = sum^{51}_{k=1}sum^{k-1}_{r=0}binom{51}{51-k}binom{50}{50-r}\
                      =& sum^{50}_{k'=0} sum^{50}_{r'=k'}binom{51}{k'}binom{50}{r'} = sum^{51}_{k'=0} sum^{50}_{r'=k'}binom{51}{k'}binom{50}{r'}.
                      end{align}
                      Hence it follows
                      begin{align}
                      2S= sum^{51}_{k=0}sum_{r=0}^{50}binom{51}{k}binom{50}{r}= sum^{51}_{k=0}binom{51}{k}sum_{r=0}^{50}binom{50}{r}=2^{51}cdot 2^{50}
                      end{align}
                      which means
                      begin{align}
                      S = 2^{100}.
                      end{align}
                      Hence you arrive at $S/2^{101} = 1/2$.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Here's an alternative solution.



                        Observe
                        begin{align}
                        S=& sum^{51}_{k=1}sum^{k-1}_{r=0}binom{51}{k}binom{50}{r} = sum^{51}_{k=1}sum^{k-1}_{r=0}binom{51}{51-k}binom{50}{50-r}\
                        =& sum^{50}_{k'=0} sum^{50}_{r'=k'}binom{51}{k'}binom{50}{r'} = sum^{51}_{k'=0} sum^{50}_{r'=k'}binom{51}{k'}binom{50}{r'}.
                        end{align}
                        Hence it follows
                        begin{align}
                        2S= sum^{51}_{k=0}sum_{r=0}^{50}binom{51}{k}binom{50}{r}= sum^{51}_{k=0}binom{51}{k}sum_{r=0}^{50}binom{50}{r}=2^{51}cdot 2^{50}
                        end{align}
                        which means
                        begin{align}
                        S = 2^{100}.
                        end{align}
                        Hence you arrive at $S/2^{101} = 1/2$.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Here's an alternative solution.



                          Observe
                          begin{align}
                          S=& sum^{51}_{k=1}sum^{k-1}_{r=0}binom{51}{k}binom{50}{r} = sum^{51}_{k=1}sum^{k-1}_{r=0}binom{51}{51-k}binom{50}{50-r}\
                          =& sum^{50}_{k'=0} sum^{50}_{r'=k'}binom{51}{k'}binom{50}{r'} = sum^{51}_{k'=0} sum^{50}_{r'=k'}binom{51}{k'}binom{50}{r'}.
                          end{align}
                          Hence it follows
                          begin{align}
                          2S= sum^{51}_{k=0}sum_{r=0}^{50}binom{51}{k}binom{50}{r}= sum^{51}_{k=0}binom{51}{k}sum_{r=0}^{50}binom{50}{r}=2^{51}cdot 2^{50}
                          end{align}
                          which means
                          begin{align}
                          S = 2^{100}.
                          end{align}
                          Hence you arrive at $S/2^{101} = 1/2$.






                          share|cite|improve this answer









                          $endgroup$



                          Here's an alternative solution.



                          Observe
                          begin{align}
                          S=& sum^{51}_{k=1}sum^{k-1}_{r=0}binom{51}{k}binom{50}{r} = sum^{51}_{k=1}sum^{k-1}_{r=0}binom{51}{51-k}binom{50}{50-r}\
                          =& sum^{50}_{k'=0} sum^{50}_{r'=k'}binom{51}{k'}binom{50}{r'} = sum^{51}_{k'=0} sum^{50}_{r'=k'}binom{51}{k'}binom{50}{r'}.
                          end{align}
                          Hence it follows
                          begin{align}
                          2S= sum^{51}_{k=0}sum_{r=0}^{50}binom{51}{k}binom{50}{r}= sum^{51}_{k=0}binom{51}{k}sum_{r=0}^{50}binom{50}{r}=2^{51}cdot 2^{50}
                          end{align}
                          which means
                          begin{align}
                          S = 2^{100}.
                          end{align}
                          Hence you arrive at $S/2^{101} = 1/2$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 16 '18 at 7:06









                          Jacky ChongJacky Chong

                          19.1k21129




                          19.1k21129






























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