An equality with factorials [duplicate]

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  • How to simplify the summation kk! without using induction?

    6 answers



Prove that
$$2times 2! + 3times 3!+4times 4! +....+ntimes n!=(n+1)!-2$$
I know that it can be proved by mathematical induction, but I want to prove it without using the mathematical induction.
I tied the equation $$C^n_0 + C^n_1 + C^n_2 +...+C^n_n=2^n$$
But I did not get any thing useful.






combinatorics







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marked as duplicate by Leucippus, Mike Earnest, stressed out, Misha Lavrov, Xander Henderson 22 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 5




    Hint: Telescoping sum.
    – user202729
    2 days ago










  • Are you sure that you have the supposed identity correct? On the left side you have several positive terms being added to $ntimes n!$, which is strictly greater than $n!$ which is strictly greater than $(n-1)!$ which is greater still than $(n-1)!-2$...
    – JMoravitz
    2 days ago






  • 1




    RHS should be $(ncolorred+1)!-2$ instead.
    – user202729
    2 days ago











  • I am sorry it is as you said that is a misstyping @JMoravitz
    – Hussien Mohamed
    2 days ago














up vote
8
down vote

favorite













This question already has an answer here:



  • How to simplify the summation kk! without using induction?

    6 answers



Prove that
$$2times 2! + 3times 3!+4times 4! +....+ntimes n!=(n+1)!-2$$
I know that it can be proved by mathematical induction, but I want to prove it without using the mathematical induction.
I tied the equation $$C^n_0 + C^n_1 + C^n_2 +...+C^n_n=2^n$$
But I did not get any thing useful.






combinatorics







share|cite|improve this question















marked as duplicate by Leucippus, Mike Earnest, stressed out, Misha Lavrov, Xander Henderson 22 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 5




    Hint: Telescoping sum.
    – user202729
    2 days ago










  • Are you sure that you have the supposed identity correct? On the left side you have several positive terms being added to $ntimes n!$, which is strictly greater than $n!$ which is strictly greater than $(n-1)!$ which is greater still than $(n-1)!-2$...
    – JMoravitz
    2 days ago






  • 1




    RHS should be $(ncolorred+1)!-2$ instead.
    – user202729
    2 days ago











  • I am sorry it is as you said that is a misstyping @JMoravitz
    – Hussien Mohamed
    2 days ago












up vote
8
down vote

favorite









up vote
8
down vote

favorite












This question already has an answer here:



  • How to simplify the summation kk! without using induction?

    6 answers



Prove that
$$2times 2! + 3times 3!+4times 4! +....+ntimes n!=(n+1)!-2$$
I know that it can be proved by mathematical induction, but I want to prove it without using the mathematical induction.
I tied the equation $$C^n_0 + C^n_1 + C^n_2 +...+C^n_n=2^n$$
But I did not get any thing useful.






combinatorics







share|cite|improve this question
















This question already has an answer here:



  • How to simplify the summation kk! without using induction?

    6 answers



Prove that
$$2times 2! + 3times 3!+4times 4! +....+ntimes n!=(n+1)!-2$$
I know that it can be proved by mathematical induction, but I want to prove it without using the mathematical induction.
I tied the equation $$C^n_0 + C^n_1 + C^n_2 +...+C^n_n=2^n$$
But I did not get any thing useful.





This question already has an answer here:



  • How to simplify the summation kk! without using induction?

    6 answers






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









N. F. Taussig

39.2k93153




39.2k93153










asked 2 days ago









Hussien Mohamed

726112




726112




marked as duplicate by Leucippus, Mike Earnest, stressed out, Misha Lavrov, Xander Henderson 22 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Leucippus, Mike Earnest, stressed out, Misha Lavrov, Xander Henderson 22 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









  • 5




    Hint: Telescoping sum.
    – user202729
    2 days ago










  • Are you sure that you have the supposed identity correct? On the left side you have several positive terms being added to $ntimes n!$, which is strictly greater than $n!$ which is strictly greater than $(n-1)!$ which is greater still than $(n-1)!-2$...
    – JMoravitz
    2 days ago






  • 1




    RHS should be $(ncolorred+1)!-2$ instead.
    – user202729
    2 days ago











  • I am sorry it is as you said that is a misstyping @JMoravitz
    – Hussien Mohamed
    2 days ago












  • 5




    Hint: Telescoping sum.
    – user202729
    2 days ago










  • Are you sure that you have the supposed identity correct? On the left side you have several positive terms being added to $ntimes n!$, which is strictly greater than $n!$ which is strictly greater than $(n-1)!$ which is greater still than $(n-1)!-2$...
    – JMoravitz
    2 days ago






  • 1




    RHS should be $(ncolorred+1)!-2$ instead.
    – user202729
    2 days ago











  • I am sorry it is as you said that is a misstyping @JMoravitz
    – Hussien Mohamed
    2 days ago







5




5




Hint: Telescoping sum.
– user202729
2 days ago




Hint: Telescoping sum.
– user202729
2 days ago












Are you sure that you have the supposed identity correct? On the left side you have several positive terms being added to $ntimes n!$, which is strictly greater than $n!$ which is strictly greater than $(n-1)!$ which is greater still than $(n-1)!-2$...
– JMoravitz
2 days ago




Are you sure that you have the supposed identity correct? On the left side you have several positive terms being added to $ntimes n!$, which is strictly greater than $n!$ which is strictly greater than $(n-1)!$ which is greater still than $(n-1)!-2$...
– JMoravitz
2 days ago




1




1




RHS should be $(ncolorred+1)!-2$ instead.
– user202729
2 days ago





RHS should be $(ncolorred+1)!-2$ instead.
– user202729
2 days ago













I am sorry it is as you said that is a misstyping @JMoravitz
– Hussien Mohamed
2 days ago




I am sorry it is as you said that is a misstyping @JMoravitz
– Hussien Mohamed
2 days ago










2 Answers
2






active

oldest

votes

















up vote
16
down vote



accepted










Hint:
$$sum_k=2^n kcdot k!=sum_k=2^n (k+1-1)cdot k!=sum_k=2^n (k+1)!-sum_k=2^n k!$$






share|cite|improve this answer



























    up vote
    14
    down vote













    $2times2!=3!-2!$



    $3times3!=4!-3!$



    ...



    $ntimes n!=(n+1)!-n!$



    Add them all up then $3!,4!,...n!$ cancel out.



    Therefore LHS$=(n+1)!-2$






    share|cite|improve this answer




















    • thank you so much @abc
      – Hussien Mohamed
      2 days ago










    • You are welcome:)
      – abc...
      2 days ago










    • does it work for any natural number or for $ n geq 2$ only @abc
      – Hussien Mohamed
      yesterday











    • @HussienMohamed the factorial function is only defined for non-negative integers.
      – abc...
      yesterday

















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    16
    down vote



    accepted










    Hint:
    $$sum_k=2^n kcdot k!=sum_k=2^n (k+1-1)cdot k!=sum_k=2^n (k+1)!-sum_k=2^n k!$$






    share|cite|improve this answer
























      up vote
      16
      down vote



      accepted










      Hint:
      $$sum_k=2^n kcdot k!=sum_k=2^n (k+1-1)cdot k!=sum_k=2^n (k+1)!-sum_k=2^n k!$$






      share|cite|improve this answer






















        up vote
        16
        down vote



        accepted







        up vote
        16
        down vote



        accepted






        Hint:
        $$sum_k=2^n kcdot k!=sum_k=2^n (k+1-1)cdot k!=sum_k=2^n (k+1)!-sum_k=2^n k!$$






        share|cite|improve this answer












        Hint:
        $$sum_k=2^n kcdot k!=sum_k=2^n (k+1-1)cdot k!=sum_k=2^n (k+1)!-sum_k=2^n k!$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        farruhota

        15.1k2734




        15.1k2734




















            up vote
            14
            down vote













            $2times2!=3!-2!$



            $3times3!=4!-3!$



            ...



            $ntimes n!=(n+1)!-n!$



            Add them all up then $3!,4!,...n!$ cancel out.



            Therefore LHS$=(n+1)!-2$






            share|cite|improve this answer




















            • thank you so much @abc
              – Hussien Mohamed
              2 days ago










            • You are welcome:)
              – abc...
              2 days ago










            • does it work for any natural number or for $ n geq 2$ only @abc
              – Hussien Mohamed
              yesterday











            • @HussienMohamed the factorial function is only defined for non-negative integers.
              – abc...
              yesterday














            up vote
            14
            down vote













            $2times2!=3!-2!$



            $3times3!=4!-3!$



            ...



            $ntimes n!=(n+1)!-n!$



            Add them all up then $3!,4!,...n!$ cancel out.



            Therefore LHS$=(n+1)!-2$






            share|cite|improve this answer




















            • thank you so much @abc
              – Hussien Mohamed
              2 days ago










            • You are welcome:)
              – abc...
              2 days ago










            • does it work for any natural number or for $ n geq 2$ only @abc
              – Hussien Mohamed
              yesterday











            • @HussienMohamed the factorial function is only defined for non-negative integers.
              – abc...
              yesterday












            up vote
            14
            down vote










            up vote
            14
            down vote









            $2times2!=3!-2!$



            $3times3!=4!-3!$



            ...



            $ntimes n!=(n+1)!-n!$



            Add them all up then $3!,4!,...n!$ cancel out.



            Therefore LHS$=(n+1)!-2$






            share|cite|improve this answer












            $2times2!=3!-2!$



            $3times3!=4!-3!$



            ...



            $ntimes n!=(n+1)!-n!$



            Add them all up then $3!,4!,...n!$ cancel out.



            Therefore LHS$=(n+1)!-2$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            abc...

            1,639527




            1,639527











            • thank you so much @abc
              – Hussien Mohamed
              2 days ago










            • You are welcome:)
              – abc...
              2 days ago










            • does it work for any natural number or for $ n geq 2$ only @abc
              – Hussien Mohamed
              yesterday











            • @HussienMohamed the factorial function is only defined for non-negative integers.
              – abc...
              yesterday
















            • thank you so much @abc
              – Hussien Mohamed
              2 days ago










            • You are welcome:)
              – abc...
              2 days ago










            • does it work for any natural number or for $ n geq 2$ only @abc
              – Hussien Mohamed
              yesterday











            • @HussienMohamed the factorial function is only defined for non-negative integers.
              – abc...
              yesterday















            thank you so much @abc
            – Hussien Mohamed
            2 days ago




            thank you so much @abc
            – Hussien Mohamed
            2 days ago












            You are welcome:)
            – abc...
            2 days ago




            You are welcome:)
            – abc...
            2 days ago












            does it work for any natural number or for $ n geq 2$ only @abc
            – Hussien Mohamed
            yesterday





            does it work for any natural number or for $ n geq 2$ only @abc
            – Hussien Mohamed
            yesterday













            @HussienMohamed the factorial function is only defined for non-negative integers.
            – abc...
            yesterday




            @HussienMohamed the factorial function is only defined for non-negative integers.
            – abc...
            yesterday


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