What will be the value of this binomial sum?

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How to evaluate this binomial sum?
$$sum_r=1^nrn-1 choose r-1(k-1)^r-1$$










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    How to evaluate this binomial sum?
    $$sum_r=1^nrn-1 choose r-1(k-1)^r-1$$










    share|cite|improve this question























      up vote
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      down vote

      favorite









      up vote
      3
      down vote

      favorite











      How to evaluate this binomial sum?
      $$sum_r=1^nrn-1 choose r-1(k-1)^r-1$$










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      How to evaluate this binomial sum?
      $$sum_r=1^nrn-1 choose r-1(k-1)^r-1$$







      combinatorics summation binomial-coefficients






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      asked 3 hours ago









      Debabrata

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      1235




















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          Hint: let $x=k-1$ begineqnarraysum_r=1^nrn-1 choose r-1x^r-1&=&Big(sum_r=1^nn-1 choose r-1x^r Big)'\
          &=&Big(xunderbracesum_r=1^nn-1 choose r-1x^r-1 Big)'\
          &=&Big(x(1+x)^n-1 Big)'\
          endeqnarray






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            Hint:



            $$rbinomn-1r-1=binomn-1r-1+(r-1)binomn-1r-1$$



            $$=binomn-1r-1+(n-1)cdotdfrac(n-2)!(r-2)!(n-2)-(r-2)$$



            $$=binomn-1r-1+(n-1)binomn-2r-2$$



            Now put $x=y=1$ in $$(x+y)^m=sum_r=0^mbinom mr x^m-ry^r$$



            Finally put $m=n-1$ and $m=n-2$






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              2 Answers
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              active

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              2 Answers
              2






              active

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              active

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              active

              oldest

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              up vote
              3
              down vote



              accepted










              Hint: let $x=k-1$ begineqnarraysum_r=1^nrn-1 choose r-1x^r-1&=&Big(sum_r=1^nn-1 choose r-1x^r Big)'\
              &=&Big(xunderbracesum_r=1^nn-1 choose r-1x^r-1 Big)'\
              &=&Big(x(1+x)^n-1 Big)'\
              endeqnarray






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                up vote
                3
                down vote



                accepted










                Hint: let $x=k-1$ begineqnarraysum_r=1^nrn-1 choose r-1x^r-1&=&Big(sum_r=1^nn-1 choose r-1x^r Big)'\
                &=&Big(xunderbracesum_r=1^nn-1 choose r-1x^r-1 Big)'\
                &=&Big(x(1+x)^n-1 Big)'\
                endeqnarray






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                  up vote
                  3
                  down vote



                  accepted







                  up vote
                  3
                  down vote



                  accepted






                  Hint: let $x=k-1$ begineqnarraysum_r=1^nrn-1 choose r-1x^r-1&=&Big(sum_r=1^nn-1 choose r-1x^r Big)'\
                  &=&Big(xunderbracesum_r=1^nn-1 choose r-1x^r-1 Big)'\
                  &=&Big(x(1+x)^n-1 Big)'\
                  endeqnarray






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                  Hint: let $x=k-1$ begineqnarraysum_r=1^nrn-1 choose r-1x^r-1&=&Big(sum_r=1^nn-1 choose r-1x^r Big)'\
                  &=&Big(xunderbracesum_r=1^nn-1 choose r-1x^r-1 Big)'\
                  &=&Big(x(1+x)^n-1 Big)'\
                  endeqnarray







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                  edited 3 hours ago

























                  answered 3 hours ago









                  greedoid

                  32k114287




                  32k114287




















                      up vote
                      1
                      down vote













                      Hint:



                      $$rbinomn-1r-1=binomn-1r-1+(r-1)binomn-1r-1$$



                      $$=binomn-1r-1+(n-1)cdotdfrac(n-2)!(r-2)!(n-2)-(r-2)$$



                      $$=binomn-1r-1+(n-1)binomn-2r-2$$



                      Now put $x=y=1$ in $$(x+y)^m=sum_r=0^mbinom mr x^m-ry^r$$



                      Finally put $m=n-1$ and $m=n-2$






                      share|cite|improve this answer
























                        up vote
                        1
                        down vote













                        Hint:



                        $$rbinomn-1r-1=binomn-1r-1+(r-1)binomn-1r-1$$



                        $$=binomn-1r-1+(n-1)cdotdfrac(n-2)!(r-2)!(n-2)-(r-2)$$



                        $$=binomn-1r-1+(n-1)binomn-2r-2$$



                        Now put $x=y=1$ in $$(x+y)^m=sum_r=0^mbinom mr x^m-ry^r$$



                        Finally put $m=n-1$ and $m=n-2$






                        share|cite|improve this answer






















                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          Hint:



                          $$rbinomn-1r-1=binomn-1r-1+(r-1)binomn-1r-1$$



                          $$=binomn-1r-1+(n-1)cdotdfrac(n-2)!(r-2)!(n-2)-(r-2)$$



                          $$=binomn-1r-1+(n-1)binomn-2r-2$$



                          Now put $x=y=1$ in $$(x+y)^m=sum_r=0^mbinom mr x^m-ry^r$$



                          Finally put $m=n-1$ and $m=n-2$






                          share|cite|improve this answer












                          Hint:



                          $$rbinomn-1r-1=binomn-1r-1+(r-1)binomn-1r-1$$



                          $$=binomn-1r-1+(n-1)cdotdfrac(n-2)!(r-2)!(n-2)-(r-2)$$



                          $$=binomn-1r-1+(n-1)binomn-2r-2$$



                          Now put $x=y=1$ in $$(x+y)^m=sum_r=0^mbinom mr x^m-ry^r$$



                          Finally put $m=n-1$ and $m=n-2$







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                          answered 2 hours ago









                          lab bhattacharjee

                          217k14153268




                          217k14153268



























                               

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