Induction Proof for Sequences












2












$begingroup$


Given a sequence $s_k=s_{k-1}+6k$, where $s_0=7$.



Question: First, find the closed formula for the $n$-th component of this sequence by hand and then prove that your formula is correct



My attempt:
I found the first couple of terms of the sequence to be
$s_0=7$,
$s_1=13$,
$s_2=25$,
$s_3=43$,
$s_4=67$ and
$s_5=97$.



I found the formula for the $n$-th term to be $s_n=3n^2+3n+7$.



Proof:
Base case $s_0=7$ therefore
$7=3cdot(0)^2+3cdot(0)+7$ so the formula works for the $s_0$ element.



I'm not sure how to proceed from here but I believe the proof should be a proof by Strong Induction. Any help will be greatly appreciated.










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    2












    $begingroup$


    Given a sequence $s_k=s_{k-1}+6k$, where $s_0=7$.



    Question: First, find the closed formula for the $n$-th component of this sequence by hand and then prove that your formula is correct



    My attempt:
    I found the first couple of terms of the sequence to be
    $s_0=7$,
    $s_1=13$,
    $s_2=25$,
    $s_3=43$,
    $s_4=67$ and
    $s_5=97$.



    I found the formula for the $n$-th term to be $s_n=3n^2+3n+7$.



    Proof:
    Base case $s_0=7$ therefore
    $7=3cdot(0)^2+3cdot(0)+7$ so the formula works for the $s_0$ element.



    I'm not sure how to proceed from here but I believe the proof should be a proof by Strong Induction. Any help will be greatly appreciated.










    share|cite|improve this question









    New contributor




    Julia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Given a sequence $s_k=s_{k-1}+6k$, where $s_0=7$.



      Question: First, find the closed formula for the $n$-th component of this sequence by hand and then prove that your formula is correct



      My attempt:
      I found the first couple of terms of the sequence to be
      $s_0=7$,
      $s_1=13$,
      $s_2=25$,
      $s_3=43$,
      $s_4=67$ and
      $s_5=97$.



      I found the formula for the $n$-th term to be $s_n=3n^2+3n+7$.



      Proof:
      Base case $s_0=7$ therefore
      $7=3cdot(0)^2+3cdot(0)+7$ so the formula works for the $s_0$ element.



      I'm not sure how to proceed from here but I believe the proof should be a proof by Strong Induction. Any help will be greatly appreciated.










      share|cite|improve this question









      New contributor




      Julia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Given a sequence $s_k=s_{k-1}+6k$, where $s_0=7$.



      Question: First, find the closed formula for the $n$-th component of this sequence by hand and then prove that your formula is correct



      My attempt:
      I found the first couple of terms of the sequence to be
      $s_0=7$,
      $s_1=13$,
      $s_2=25$,
      $s_3=43$,
      $s_4=67$ and
      $s_5=97$.



      I found the formula for the $n$-th term to be $s_n=3n^2+3n+7$.



      Proof:
      Base case $s_0=7$ therefore
      $7=3cdot(0)^2+3cdot(0)+7$ so the formula works for the $s_0$ element.



      I'm not sure how to proceed from here but I believe the proof should be a proof by Strong Induction. Any help will be greatly appreciated.







      sequences-and-series induction






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      Julia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      edited 4 hours ago









      Marian G.

      38426




      38426






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      asked 4 hours ago









      JuliaJulia

      113




      113




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          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          You just apply the recurrence
          to your induction hypothesis.



          If
          $s_n=3n^2+3n+7
          $
          ,
          then,
          since
          $s_k=s_{k-1}+6k
          $
          ,



          $begin{array}\
          s_{n+1}
          &=s_n+6(n+1)\
          &=3n^2+3n+7+6(n+1)\
          &=3n^2+9n+13\
          end{array}
          $



          If your hypothesis is true,
          then



          $begin{array}\
          s_{n+1}
          &=3(n+1)^2+3(n+1)+7\
          &=3(n^2+2n+1)+3(n+1)+7\
          &=3n^2+6n+3+3n+3+7\
          &=3n^2+9n+13\
          end{array}
          $



          This matches the result
          from the induction step,
          so the induction hypothesis is proved.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            FYI, here is a way to determine what the closed formula is without having to determine it by hand. This is an extension of the answer given by szw1710. The given sequence is



            $$s_k=s_{k-1}+6k tag{1}label{eq1}$$



            Summing both sides of eqref{eq1} from $1$ to $n$ gives that



            $sum_{k=1}^{n}s_k = sum_{k=1}^{n}s_{k-1} + sum_{k=1}^{n}6k$



            $sum_{k=1}^{n-1}s_k + s_n = s_0 + sum_{k=1}^{n-1}s_{k} + 6frac{n(n+1)}{2}$



            $s_n = 7 + 3n(n+1) = 3n^2 + 3n + 7$



            As you can see, this technique can easily be used in any cases where you have $s_k = s_{k-1} + f(k)$ and the sum of $f(k)$ up to $k = n$ can be fairly easily determined.



            More generally, your question is a fairly simple example of Linear Recurrence Relations with Constant Coefficients. You can use a certain technique of a characteristic equation, as described in that link, to directly determine the solution of even considerably more complicated such equations.






            share|cite|improve this answer











            $endgroup$





















              2












              $begingroup$

              It seems to me that induction is not needed here. Fix $kinBbb N.$ A direct computation shows that $s_k-s_{k-1}=6k$ and $s_0=7.$ However, there is another problem: both induction, and my method show that if $s_k=3k^2+3k+7$, then $s_{k}=s_{k-1}+6k$. In fact, whe should prove the converse: if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $s_k=3k^2+3k+7$. I will look for some reasoning going in this direction.



              Let $s_0=7$ and $s_{k}=s_{k-1}+6k.$ Define $t_k=s_k-3k^2-3k-7.$ Then $t_0=0$. It is easy to prove (by direct computation) that $t_k=t_{k-1}$, so $(t_k)$ is a constant (in fact, zero) sequence. Then $s_k=3k^2+3k+7.$






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                My answer has one way to show what you're asking about, i.e., "if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $a_k=3k^2+3k+7$".
                $endgroup$
                – John Omielan
                2 hours ago










              • $begingroup$
                Yes, of course. About such techniques one could read in the "Concrete mathematics" book by Graham, Knuth, Patashnik.
                $endgroup$
                – szw1710
                2 hours ago












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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              You just apply the recurrence
              to your induction hypothesis.



              If
              $s_n=3n^2+3n+7
              $
              ,
              then,
              since
              $s_k=s_{k-1}+6k
              $
              ,



              $begin{array}\
              s_{n+1}
              &=s_n+6(n+1)\
              &=3n^2+3n+7+6(n+1)\
              &=3n^2+9n+13\
              end{array}
              $



              If your hypothesis is true,
              then



              $begin{array}\
              s_{n+1}
              &=3(n+1)^2+3(n+1)+7\
              &=3(n^2+2n+1)+3(n+1)+7\
              &=3n^2+6n+3+3n+3+7\
              &=3n^2+9n+13\
              end{array}
              $



              This matches the result
              from the induction step,
              so the induction hypothesis is proved.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                You just apply the recurrence
                to your induction hypothesis.



                If
                $s_n=3n^2+3n+7
                $
                ,
                then,
                since
                $s_k=s_{k-1}+6k
                $
                ,



                $begin{array}\
                s_{n+1}
                &=s_n+6(n+1)\
                &=3n^2+3n+7+6(n+1)\
                &=3n^2+9n+13\
                end{array}
                $



                If your hypothesis is true,
                then



                $begin{array}\
                s_{n+1}
                &=3(n+1)^2+3(n+1)+7\
                &=3(n^2+2n+1)+3(n+1)+7\
                &=3n^2+6n+3+3n+3+7\
                &=3n^2+9n+13\
                end{array}
                $



                This matches the result
                from the induction step,
                so the induction hypothesis is proved.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  You just apply the recurrence
                  to your induction hypothesis.



                  If
                  $s_n=3n^2+3n+7
                  $
                  ,
                  then,
                  since
                  $s_k=s_{k-1}+6k
                  $
                  ,



                  $begin{array}\
                  s_{n+1}
                  &=s_n+6(n+1)\
                  &=3n^2+3n+7+6(n+1)\
                  &=3n^2+9n+13\
                  end{array}
                  $



                  If your hypothesis is true,
                  then



                  $begin{array}\
                  s_{n+1}
                  &=3(n+1)^2+3(n+1)+7\
                  &=3(n^2+2n+1)+3(n+1)+7\
                  &=3n^2+6n+3+3n+3+7\
                  &=3n^2+9n+13\
                  end{array}
                  $



                  This matches the result
                  from the induction step,
                  so the induction hypothesis is proved.






                  share|cite|improve this answer









                  $endgroup$



                  You just apply the recurrence
                  to your induction hypothesis.



                  If
                  $s_n=3n^2+3n+7
                  $
                  ,
                  then,
                  since
                  $s_k=s_{k-1}+6k
                  $
                  ,



                  $begin{array}\
                  s_{n+1}
                  &=s_n+6(n+1)\
                  &=3n^2+3n+7+6(n+1)\
                  &=3n^2+9n+13\
                  end{array}
                  $



                  If your hypothesis is true,
                  then



                  $begin{array}\
                  s_{n+1}
                  &=3(n+1)^2+3(n+1)+7\
                  &=3(n^2+2n+1)+3(n+1)+7\
                  &=3n^2+6n+3+3n+3+7\
                  &=3n^2+9n+13\
                  end{array}
                  $



                  This matches the result
                  from the induction step,
                  so the induction hypothesis is proved.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 4 hours ago









                  marty cohenmarty cohen

                  75.9k549130




                  75.9k549130























                      2












                      $begingroup$

                      FYI, here is a way to determine what the closed formula is without having to determine it by hand. This is an extension of the answer given by szw1710. The given sequence is



                      $$s_k=s_{k-1}+6k tag{1}label{eq1}$$



                      Summing both sides of eqref{eq1} from $1$ to $n$ gives that



                      $sum_{k=1}^{n}s_k = sum_{k=1}^{n}s_{k-1} + sum_{k=1}^{n}6k$



                      $sum_{k=1}^{n-1}s_k + s_n = s_0 + sum_{k=1}^{n-1}s_{k} + 6frac{n(n+1)}{2}$



                      $s_n = 7 + 3n(n+1) = 3n^2 + 3n + 7$



                      As you can see, this technique can easily be used in any cases where you have $s_k = s_{k-1} + f(k)$ and the sum of $f(k)$ up to $k = n$ can be fairly easily determined.



                      More generally, your question is a fairly simple example of Linear Recurrence Relations with Constant Coefficients. You can use a certain technique of a characteristic equation, as described in that link, to directly determine the solution of even considerably more complicated such equations.






                      share|cite|improve this answer











                      $endgroup$


















                        2












                        $begingroup$

                        FYI, here is a way to determine what the closed formula is without having to determine it by hand. This is an extension of the answer given by szw1710. The given sequence is



                        $$s_k=s_{k-1}+6k tag{1}label{eq1}$$



                        Summing both sides of eqref{eq1} from $1$ to $n$ gives that



                        $sum_{k=1}^{n}s_k = sum_{k=1}^{n}s_{k-1} + sum_{k=1}^{n}6k$



                        $sum_{k=1}^{n-1}s_k + s_n = s_0 + sum_{k=1}^{n-1}s_{k} + 6frac{n(n+1)}{2}$



                        $s_n = 7 + 3n(n+1) = 3n^2 + 3n + 7$



                        As you can see, this technique can easily be used in any cases where you have $s_k = s_{k-1} + f(k)$ and the sum of $f(k)$ up to $k = n$ can be fairly easily determined.



                        More generally, your question is a fairly simple example of Linear Recurrence Relations with Constant Coefficients. You can use a certain technique of a characteristic equation, as described in that link, to directly determine the solution of even considerably more complicated such equations.






                        share|cite|improve this answer











                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          FYI, here is a way to determine what the closed formula is without having to determine it by hand. This is an extension of the answer given by szw1710. The given sequence is



                          $$s_k=s_{k-1}+6k tag{1}label{eq1}$$



                          Summing both sides of eqref{eq1} from $1$ to $n$ gives that



                          $sum_{k=1}^{n}s_k = sum_{k=1}^{n}s_{k-1} + sum_{k=1}^{n}6k$



                          $sum_{k=1}^{n-1}s_k + s_n = s_0 + sum_{k=1}^{n-1}s_{k} + 6frac{n(n+1)}{2}$



                          $s_n = 7 + 3n(n+1) = 3n^2 + 3n + 7$



                          As you can see, this technique can easily be used in any cases where you have $s_k = s_{k-1} + f(k)$ and the sum of $f(k)$ up to $k = n$ can be fairly easily determined.



                          More generally, your question is a fairly simple example of Linear Recurrence Relations with Constant Coefficients. You can use a certain technique of a characteristic equation, as described in that link, to directly determine the solution of even considerably more complicated such equations.






                          share|cite|improve this answer











                          $endgroup$



                          FYI, here is a way to determine what the closed formula is without having to determine it by hand. This is an extension of the answer given by szw1710. The given sequence is



                          $$s_k=s_{k-1}+6k tag{1}label{eq1}$$



                          Summing both sides of eqref{eq1} from $1$ to $n$ gives that



                          $sum_{k=1}^{n}s_k = sum_{k=1}^{n}s_{k-1} + sum_{k=1}^{n}6k$



                          $sum_{k=1}^{n-1}s_k + s_n = s_0 + sum_{k=1}^{n-1}s_{k} + 6frac{n(n+1)}{2}$



                          $s_n = 7 + 3n(n+1) = 3n^2 + 3n + 7$



                          As you can see, this technique can easily be used in any cases where you have $s_k = s_{k-1} + f(k)$ and the sum of $f(k)$ up to $k = n$ can be fairly easily determined.



                          More generally, your question is a fairly simple example of Linear Recurrence Relations with Constant Coefficients. You can use a certain technique of a characteristic equation, as described in that link, to directly determine the solution of even considerably more complicated such equations.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 2 hours ago

























                          answered 2 hours ago









                          John OmielanJohn Omielan

                          5,2192218




                          5,2192218























                              2












                              $begingroup$

                              It seems to me that induction is not needed here. Fix $kinBbb N.$ A direct computation shows that $s_k-s_{k-1}=6k$ and $s_0=7.$ However, there is another problem: both induction, and my method show that if $s_k=3k^2+3k+7$, then $s_{k}=s_{k-1}+6k$. In fact, whe should prove the converse: if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $s_k=3k^2+3k+7$. I will look for some reasoning going in this direction.



                              Let $s_0=7$ and $s_{k}=s_{k-1}+6k.$ Define $t_k=s_k-3k^2-3k-7.$ Then $t_0=0$. It is easy to prove (by direct computation) that $t_k=t_{k-1}$, so $(t_k)$ is a constant (in fact, zero) sequence. Then $s_k=3k^2+3k+7.$






                              share|cite|improve this answer











                              $endgroup$









                              • 1




                                $begingroup$
                                My answer has one way to show what you're asking about, i.e., "if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $a_k=3k^2+3k+7$".
                                $endgroup$
                                – John Omielan
                                2 hours ago










                              • $begingroup$
                                Yes, of course. About such techniques one could read in the "Concrete mathematics" book by Graham, Knuth, Patashnik.
                                $endgroup$
                                – szw1710
                                2 hours ago
















                              2












                              $begingroup$

                              It seems to me that induction is not needed here. Fix $kinBbb N.$ A direct computation shows that $s_k-s_{k-1}=6k$ and $s_0=7.$ However, there is another problem: both induction, and my method show that if $s_k=3k^2+3k+7$, then $s_{k}=s_{k-1}+6k$. In fact, whe should prove the converse: if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $s_k=3k^2+3k+7$. I will look for some reasoning going in this direction.



                              Let $s_0=7$ and $s_{k}=s_{k-1}+6k.$ Define $t_k=s_k-3k^2-3k-7.$ Then $t_0=0$. It is easy to prove (by direct computation) that $t_k=t_{k-1}$, so $(t_k)$ is a constant (in fact, zero) sequence. Then $s_k=3k^2+3k+7.$






                              share|cite|improve this answer











                              $endgroup$









                              • 1




                                $begingroup$
                                My answer has one way to show what you're asking about, i.e., "if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $a_k=3k^2+3k+7$".
                                $endgroup$
                                – John Omielan
                                2 hours ago










                              • $begingroup$
                                Yes, of course. About such techniques one could read in the "Concrete mathematics" book by Graham, Knuth, Patashnik.
                                $endgroup$
                                – szw1710
                                2 hours ago














                              2












                              2








                              2





                              $begingroup$

                              It seems to me that induction is not needed here. Fix $kinBbb N.$ A direct computation shows that $s_k-s_{k-1}=6k$ and $s_0=7.$ However, there is another problem: both induction, and my method show that if $s_k=3k^2+3k+7$, then $s_{k}=s_{k-1}+6k$. In fact, whe should prove the converse: if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $s_k=3k^2+3k+7$. I will look for some reasoning going in this direction.



                              Let $s_0=7$ and $s_{k}=s_{k-1}+6k.$ Define $t_k=s_k-3k^2-3k-7.$ Then $t_0=0$. It is easy to prove (by direct computation) that $t_k=t_{k-1}$, so $(t_k)$ is a constant (in fact, zero) sequence. Then $s_k=3k^2+3k+7.$






                              share|cite|improve this answer











                              $endgroup$



                              It seems to me that induction is not needed here. Fix $kinBbb N.$ A direct computation shows that $s_k-s_{k-1}=6k$ and $s_0=7.$ However, there is another problem: both induction, and my method show that if $s_k=3k^2+3k+7$, then $s_{k}=s_{k-1}+6k$. In fact, whe should prove the converse: if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $s_k=3k^2+3k+7$. I will look for some reasoning going in this direction.



                              Let $s_0=7$ and $s_{k}=s_{k-1}+6k.$ Define $t_k=s_k-3k^2-3k-7.$ Then $t_0=0$. It is easy to prove (by direct computation) that $t_k=t_{k-1}$, so $(t_k)$ is a constant (in fact, zero) sequence. Then $s_k=3k^2+3k+7.$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited 2 hours ago

























                              answered 4 hours ago









                              szw1710szw1710

                              6,5701223




                              6,5701223








                              • 1




                                $begingroup$
                                My answer has one way to show what you're asking about, i.e., "if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $a_k=3k^2+3k+7$".
                                $endgroup$
                                – John Omielan
                                2 hours ago










                              • $begingroup$
                                Yes, of course. About such techniques one could read in the "Concrete mathematics" book by Graham, Knuth, Patashnik.
                                $endgroup$
                                – szw1710
                                2 hours ago














                              • 1




                                $begingroup$
                                My answer has one way to show what you're asking about, i.e., "if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $a_k=3k^2+3k+7$".
                                $endgroup$
                                – John Omielan
                                2 hours ago










                              • $begingroup$
                                Yes, of course. About such techniques one could read in the "Concrete mathematics" book by Graham, Knuth, Patashnik.
                                $endgroup$
                                – szw1710
                                2 hours ago








                              1




                              1




                              $begingroup$
                              My answer has one way to show what you're asking about, i.e., "if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $a_k=3k^2+3k+7$".
                              $endgroup$
                              – John Omielan
                              2 hours ago




                              $begingroup$
                              My answer has one way to show what you're asking about, i.e., "if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $a_k=3k^2+3k+7$".
                              $endgroup$
                              – John Omielan
                              2 hours ago












                              $begingroup$
                              Yes, of course. About such techniques one could read in the "Concrete mathematics" book by Graham, Knuth, Patashnik.
                              $endgroup$
                              – szw1710
                              2 hours ago




                              $begingroup$
                              Yes, of course. About such techniques one could read in the "Concrete mathematics" book by Graham, Knuth, Patashnik.
                              $endgroup$
                              – szw1710
                              2 hours ago










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