Prove $2times1!+5times2!+10times3!+…+(n^2+1)n!=n(n+1)!$ for all positive integers












0












$begingroup$


I am trying to prove by mathematical induction $2times1!+5times2!+10times3!+...+(n^2+1)n!=n(n+1)!$ for all positive integers $n$.



So far I have:




  • Solved in the first case possible - $1$


  • Assumed the function to be true for all positive integers


  • Subbed in $k+1$ for $n$ leaving me with the following
    $$((k+1)^2+1)(k+1)!=(k+1)(k+2)!$$


I am unsure where to go from here? Any help would be greatly appreciated!






induction factorial







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$endgroup$












  • $begingroup$
    Trying to show that the equality is true
    $endgroup$
    – Holly Millican
    Dec 12 '18 at 23:26










  • $begingroup$
    Wait, the equality is false.
    $endgroup$
    – Mindlack
    Dec 12 '18 at 23:27






  • 1




    $begingroup$
    Shouldn't it be $(2 times 1!) + dots + ((k+1)^2 + 1)(k+1)!$?
    $endgroup$
    – platty
    Dec 12 '18 at 23:27










  • $begingroup$
    yes sorry that is my typo, will fix
    $endgroup$
    – Holly Millican
    Dec 12 '18 at 23:27
















0












$begingroup$


I am trying to prove by mathematical induction $2times1!+5times2!+10times3!+...+(n^2+1)n!=n(n+1)!$ for all positive integers $n$.



So far I have:




  • Solved in the first case possible - $1$


  • Assumed the function to be true for all positive integers


  • Subbed in $k+1$ for $n$ leaving me with the following
    $$((k+1)^2+1)(k+1)!=(k+1)(k+2)!$$


I am unsure where to go from here? Any help would be greatly appreciated!






induction factorial







share|cite|improve this question











$endgroup$












  • $begingroup$
    Trying to show that the equality is true
    $endgroup$
    – Holly Millican
    Dec 12 '18 at 23:26










  • $begingroup$
    Wait, the equality is false.
    $endgroup$
    – Mindlack
    Dec 12 '18 at 23:27






  • 1




    $begingroup$
    Shouldn't it be $(2 times 1!) + dots + ((k+1)^2 + 1)(k+1)!$?
    $endgroup$
    – platty
    Dec 12 '18 at 23:27










  • $begingroup$
    yes sorry that is my typo, will fix
    $endgroup$
    – Holly Millican
    Dec 12 '18 at 23:27














0












0








0


1



$begingroup$


I am trying to prove by mathematical induction $2times1!+5times2!+10times3!+...+(n^2+1)n!=n(n+1)!$ for all positive integers $n$.



So far I have:




  • Solved in the first case possible - $1$


  • Assumed the function to be true for all positive integers


  • Subbed in $k+1$ for $n$ leaving me with the following
    $$((k+1)^2+1)(k+1)!=(k+1)(k+2)!$$


I am unsure where to go from here? Any help would be greatly appreciated!






induction factorial







share|cite|improve this question











$endgroup$




I am trying to prove by mathematical induction $2times1!+5times2!+10times3!+...+(n^2+1)n!=n(n+1)!$ for all positive integers $n$.



So far I have:




  • Solved in the first case possible - $1$


  • Assumed the function to be true for all positive integers


  • Subbed in $k+1$ for $n$ leaving me with the following
    $$((k+1)^2+1)(k+1)!=(k+1)(k+2)!$$


I am unsure where to go from here? Any help would be greatly appreciated!






induction factorial




induction factorial






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 23:28







Holly Millican

















asked Dec 12 '18 at 23:25









Holly MillicanHolly Millican

528




528












  • $begingroup$
    Trying to show that the equality is true
    $endgroup$
    – Holly Millican
    Dec 12 '18 at 23:26










  • $begingroup$
    Wait, the equality is false.
    $endgroup$
    – Mindlack
    Dec 12 '18 at 23:27






  • 1




    $begingroup$
    Shouldn't it be $(2 times 1!) + dots + ((k+1)^2 + 1)(k+1)!$?
    $endgroup$
    – platty
    Dec 12 '18 at 23:27










  • $begingroup$
    yes sorry that is my typo, will fix
    $endgroup$
    – Holly Millican
    Dec 12 '18 at 23:27


















  • $begingroup$
    Trying to show that the equality is true
    $endgroup$
    – Holly Millican
    Dec 12 '18 at 23:26










  • $begingroup$
    Wait, the equality is false.
    $endgroup$
    – Mindlack
    Dec 12 '18 at 23:27






  • 1




    $begingroup$
    Shouldn't it be $(2 times 1!) + dots + ((k+1)^2 + 1)(k+1)!$?
    $endgroup$
    – platty
    Dec 12 '18 at 23:27










  • $begingroup$
    yes sorry that is my typo, will fix
    $endgroup$
    – Holly Millican
    Dec 12 '18 at 23:27
















$begingroup$
Trying to show that the equality is true
$endgroup$
– Holly Millican
Dec 12 '18 at 23:26




$begingroup$
Trying to show that the equality is true
$endgroup$
– Holly Millican
Dec 12 '18 at 23:26












$begingroup$
Wait, the equality is false.
$endgroup$
– Mindlack
Dec 12 '18 at 23:27




$begingroup$
Wait, the equality is false.
$endgroup$
– Mindlack
Dec 12 '18 at 23:27




1




1




$begingroup$
Shouldn't it be $(2 times 1!) + dots + ((k+1)^2 + 1)(k+1)!$?
$endgroup$
– platty
Dec 12 '18 at 23:27




$begingroup$
Shouldn't it be $(2 times 1!) + dots + ((k+1)^2 + 1)(k+1)!$?
$endgroup$
– platty
Dec 12 '18 at 23:27












$begingroup$
yes sorry that is my typo, will fix
$endgroup$
– Holly Millican
Dec 12 '18 at 23:27




$begingroup$
yes sorry that is my typo, will fix
$endgroup$
– Holly Millican
Dec 12 '18 at 23:27










3 Answers
3






active

oldest

votes


















0












$begingroup$

you have your base case:



Assume:



$2times1!+5times2!+10times3!+...+(n^2+1)n!=n(n+1)!$



We must show that



$2times1!+cdots+(n^2+1)n! + ((n+1)^2 + 1)(n+1)!=(n+1)(n+2)!$



$n(n+1)! + ((n+1)^2 + 1)(n+1)!$



By the inductive hypothesis.



$((n+1)^2+ n + 1)(n+1)!\
(n^2 + 3n + 2)(n+1)!\
(n+1)(n+2)(n+1)!\
(n+1)(n+2)!$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The Induction Hypothesis should give you $2 times 1! + dots + (k^2+1)k! = k(k+1)!$. In the Induction Step, you should start from one side of the equality and try to reach the other side. Here, it makes most sense to work with the summation side, as this will allow us to easily apply the Induction Hypothesis. Applying the Induction Hypothesis, we get:
    $$
    begin{align*}
    2 times 1! + dots + (k^2+1)k! + ((k+1)^2 + 1)(k+1)! &= k(k+1)! + ((k+1)^2 + 1)(k+1)! \
    &= ((k+1)^2 + k + 1) cdot (k+1)! \
    &= (k+1 + 1)(k+1) cdot (k+1)! \
    &= (k+1)(k+2)!
    end{align*}
    $$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      The key point is form the induction hypotesis



      $$2times1!+5times2!+10times3!+…+(k^2+1)k!=k(k+1)!$$



      is that



      $$2times1!+ldots+((k+1)^2+1)(k+1)!stackrel{Ind. Hyp.}=k(k+1)!+((k+1)^2+1)(k+1)!stackrel{?}=(k+1)(k+2)!$$



      and the latter is true indeed



      $$k(k+1)!+((k+1)^2+1)(k+1)!=(k+1)!(k^2+3k+2)=(k+1)!(k+1)(k+2)=(k+1)(k+2)!$$






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        0












        $begingroup$

        you have your base case:



        Assume:



        $2times1!+5times2!+10times3!+...+(n^2+1)n!=n(n+1)!$



        We must show that



        $2times1!+cdots+(n^2+1)n! + ((n+1)^2 + 1)(n+1)!=(n+1)(n+2)!$



        $n(n+1)! + ((n+1)^2 + 1)(n+1)!$



        By the inductive hypothesis.



        $((n+1)^2+ n + 1)(n+1)!\
        (n^2 + 3n + 2)(n+1)!\
        (n+1)(n+2)(n+1)!\
        (n+1)(n+2)!$






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          you have your base case:



          Assume:



          $2times1!+5times2!+10times3!+...+(n^2+1)n!=n(n+1)!$



          We must show that



          $2times1!+cdots+(n^2+1)n! + ((n+1)^2 + 1)(n+1)!=(n+1)(n+2)!$



          $n(n+1)! + ((n+1)^2 + 1)(n+1)!$



          By the inductive hypothesis.



          $((n+1)^2+ n + 1)(n+1)!\
          (n^2 + 3n + 2)(n+1)!\
          (n+1)(n+2)(n+1)!\
          (n+1)(n+2)!$






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            you have your base case:



            Assume:



            $2times1!+5times2!+10times3!+...+(n^2+1)n!=n(n+1)!$



            We must show that



            $2times1!+cdots+(n^2+1)n! + ((n+1)^2 + 1)(n+1)!=(n+1)(n+2)!$



            $n(n+1)! + ((n+1)^2 + 1)(n+1)!$



            By the inductive hypothesis.



            $((n+1)^2+ n + 1)(n+1)!\
            (n^2 + 3n + 2)(n+1)!\
            (n+1)(n+2)(n+1)!\
            (n+1)(n+2)!$






            share|cite|improve this answer









            $endgroup$



            you have your base case:



            Assume:



            $2times1!+5times2!+10times3!+...+(n^2+1)n!=n(n+1)!$



            We must show that



            $2times1!+cdots+(n^2+1)n! + ((n+1)^2 + 1)(n+1)!=(n+1)(n+2)!$



            $n(n+1)! + ((n+1)^2 + 1)(n+1)!$



            By the inductive hypothesis.



            $((n+1)^2+ n + 1)(n+1)!\
            (n^2 + 3n + 2)(n+1)!\
            (n+1)(n+2)(n+1)!\
            (n+1)(n+2)!$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 12 '18 at 23:31









            Doug MDoug M

            45.2k31954




            45.2k31954























                0












                $begingroup$

                The Induction Hypothesis should give you $2 times 1! + dots + (k^2+1)k! = k(k+1)!$. In the Induction Step, you should start from one side of the equality and try to reach the other side. Here, it makes most sense to work with the summation side, as this will allow us to easily apply the Induction Hypothesis. Applying the Induction Hypothesis, we get:
                $$
                begin{align*}
                2 times 1! + dots + (k^2+1)k! + ((k+1)^2 + 1)(k+1)! &= k(k+1)! + ((k+1)^2 + 1)(k+1)! \
                &= ((k+1)^2 + k + 1) cdot (k+1)! \
                &= (k+1 + 1)(k+1) cdot (k+1)! \
                &= (k+1)(k+2)!
                end{align*}
                $$






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  The Induction Hypothesis should give you $2 times 1! + dots + (k^2+1)k! = k(k+1)!$. In the Induction Step, you should start from one side of the equality and try to reach the other side. Here, it makes most sense to work with the summation side, as this will allow us to easily apply the Induction Hypothesis. Applying the Induction Hypothesis, we get:
                  $$
                  begin{align*}
                  2 times 1! + dots + (k^2+1)k! + ((k+1)^2 + 1)(k+1)! &= k(k+1)! + ((k+1)^2 + 1)(k+1)! \
                  &= ((k+1)^2 + k + 1) cdot (k+1)! \
                  &= (k+1 + 1)(k+1) cdot (k+1)! \
                  &= (k+1)(k+2)!
                  end{align*}
                  $$






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    The Induction Hypothesis should give you $2 times 1! + dots + (k^2+1)k! = k(k+1)!$. In the Induction Step, you should start from one side of the equality and try to reach the other side. Here, it makes most sense to work with the summation side, as this will allow us to easily apply the Induction Hypothesis. Applying the Induction Hypothesis, we get:
                    $$
                    begin{align*}
                    2 times 1! + dots + (k^2+1)k! + ((k+1)^2 + 1)(k+1)! &= k(k+1)! + ((k+1)^2 + 1)(k+1)! \
                    &= ((k+1)^2 + k + 1) cdot (k+1)! \
                    &= (k+1 + 1)(k+1) cdot (k+1)! \
                    &= (k+1)(k+2)!
                    end{align*}
                    $$






                    share|cite|improve this answer









                    $endgroup$



                    The Induction Hypothesis should give you $2 times 1! + dots + (k^2+1)k! = k(k+1)!$. In the Induction Step, you should start from one side of the equality and try to reach the other side. Here, it makes most sense to work with the summation side, as this will allow us to easily apply the Induction Hypothesis. Applying the Induction Hypothesis, we get:
                    $$
                    begin{align*}
                    2 times 1! + dots + (k^2+1)k! + ((k+1)^2 + 1)(k+1)! &= k(k+1)! + ((k+1)^2 + 1)(k+1)! \
                    &= ((k+1)^2 + k + 1) cdot (k+1)! \
                    &= (k+1 + 1)(k+1) cdot (k+1)! \
                    &= (k+1)(k+2)!
                    end{align*}
                    $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 12 '18 at 23:32









                    plattyplatty

                    3,370320




                    3,370320























                        0












                        $begingroup$

                        The key point is form the induction hypotesis



                        $$2times1!+5times2!+10times3!+…+(k^2+1)k!=k(k+1)!$$



                        is that



                        $$2times1!+ldots+((k+1)^2+1)(k+1)!stackrel{Ind. Hyp.}=k(k+1)!+((k+1)^2+1)(k+1)!stackrel{?}=(k+1)(k+2)!$$



                        and the latter is true indeed



                        $$k(k+1)!+((k+1)^2+1)(k+1)!=(k+1)!(k^2+3k+2)=(k+1)!(k+1)(k+2)=(k+1)(k+2)!$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          The key point is form the induction hypotesis



                          $$2times1!+5times2!+10times3!+…+(k^2+1)k!=k(k+1)!$$



                          is that



                          $$2times1!+ldots+((k+1)^2+1)(k+1)!stackrel{Ind. Hyp.}=k(k+1)!+((k+1)^2+1)(k+1)!stackrel{?}=(k+1)(k+2)!$$



                          and the latter is true indeed



                          $$k(k+1)!+((k+1)^2+1)(k+1)!=(k+1)!(k^2+3k+2)=(k+1)!(k+1)(k+2)=(k+1)(k+2)!$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            The key point is form the induction hypotesis



                            $$2times1!+5times2!+10times3!+…+(k^2+1)k!=k(k+1)!$$



                            is that



                            $$2times1!+ldots+((k+1)^2+1)(k+1)!stackrel{Ind. Hyp.}=k(k+1)!+((k+1)^2+1)(k+1)!stackrel{?}=(k+1)(k+2)!$$



                            and the latter is true indeed



                            $$k(k+1)!+((k+1)^2+1)(k+1)!=(k+1)!(k^2+3k+2)=(k+1)!(k+1)(k+2)=(k+1)(k+2)!$$






                            share|cite|improve this answer









                            $endgroup$



                            The key point is form the induction hypotesis



                            $$2times1!+5times2!+10times3!+…+(k^2+1)k!=k(k+1)!$$



                            is that



                            $$2times1!+ldots+((k+1)^2+1)(k+1)!stackrel{Ind. Hyp.}=k(k+1)!+((k+1)^2+1)(k+1)!stackrel{?}=(k+1)(k+2)!$$



                            and the latter is true indeed



                            $$k(k+1)!+((k+1)^2+1)(k+1)!=(k+1)!(k^2+3k+2)=(k+1)!(k+1)(k+2)=(k+1)(k+2)!$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 12 '18 at 23:52









                            gimusigimusi

                            92.9k84494




                            92.9k84494






























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