Solve recursion












2












$begingroup$


$a_n=n^2times a_{(n-1)}$, for $ n>0, a_0=1$



Whats the correct formula to use in this case?



I solved problem with recursion before, but they had '$+$' instead of $'times'$ so there is following formula to simply solve them:
$ar^n+bnr^n$.



Is there something similar, but for multiplication instead?






discrete-mathematics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Compute the first terms for $sqrt{a_n}$.
    $endgroup$
    – Mindlack
    Jan 26 at 23:29
















2












$begingroup$


$a_n=n^2times a_{(n-1)}$, for $ n>0, a_0=1$



Whats the correct formula to use in this case?



I solved problem with recursion before, but they had '$+$' instead of $'times'$ so there is following formula to simply solve them:
$ar^n+bnr^n$.



Is there something similar, but for multiplication instead?






discrete-mathematics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Compute the first terms for $sqrt{a_n}$.
    $endgroup$
    – Mindlack
    Jan 26 at 23:29














2












2








2





$begingroup$


$a_n=n^2times a_{(n-1)}$, for $ n>0, a_0=1$



Whats the correct formula to use in this case?



I solved problem with recursion before, but they had '$+$' instead of $'times'$ so there is following formula to simply solve them:
$ar^n+bnr^n$.



Is there something similar, but for multiplication instead?






discrete-mathematics







share|cite|improve this question











$endgroup$




$a_n=n^2times a_{(n-1)}$, for $ n>0, a_0=1$



Whats the correct formula to use in this case?



I solved problem with recursion before, but they had '$+$' instead of $'times'$ so there is following formula to simply solve them:
$ar^n+bnr^n$.



Is there something similar, but for multiplication instead?






discrete-mathematics




discrete-mathematics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 26 at 23:29









abc...

3,237738




3,237738










asked Jan 26 at 23:27









GorossoGorosso

315




315








  • 1




    $begingroup$
    Compute the first terms for $sqrt{a_n}$.
    $endgroup$
    – Mindlack
    Jan 26 at 23:29














  • 1




    $begingroup$
    Compute the first terms for $sqrt{a_n}$.
    $endgroup$
    – Mindlack
    Jan 26 at 23:29








1




1




$begingroup$
Compute the first terms for $sqrt{a_n}$.
$endgroup$
– Mindlack
Jan 26 at 23:29




$begingroup$
Compute the first terms for $sqrt{a_n}$.
$endgroup$
– Mindlack
Jan 26 at 23:29










2 Answers
2






active

oldest

votes


















2












$begingroup$

$a_n=(n!)^2$ using induction. Try to evaluate the first few terms on your own.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is the only reasonable answer, but it shows that the problem is somewhat stupid as $n!$ is defined literally like that, i.e. $n! = n cdot (n-1) cdot ldots cdot 2 cdot 1$...
    $endgroup$
    – Klaus
    Jan 26 at 23:37





















1












$begingroup$

Multiply these lines together:
begin{align}
a_n&=n^2times a_{n-1}\
a_{n-1}&=(n-1)^2times a_{n-2}\
&dots\
a_2&=2^2times a_{1}\
a_1&=1^2times a_{0}\
end{align}
After canceling, we get
$$
a_n = 1^2 cdot 2^2 cdots n^2 a_0 = (1 cdot 2 cdots n)^2 = (n!)^2
$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    $a_n=(n!)^2$ using induction. Try to evaluate the first few terms on your own.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This is the only reasonable answer, but it shows that the problem is somewhat stupid as $n!$ is defined literally like that, i.e. $n! = n cdot (n-1) cdot ldots cdot 2 cdot 1$...
      $endgroup$
      – Klaus
      Jan 26 at 23:37


















    2












    $begingroup$

    $a_n=(n!)^2$ using induction. Try to evaluate the first few terms on your own.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This is the only reasonable answer, but it shows that the problem is somewhat stupid as $n!$ is defined literally like that, i.e. $n! = n cdot (n-1) cdot ldots cdot 2 cdot 1$...
      $endgroup$
      – Klaus
      Jan 26 at 23:37
















    2












    2








    2





    $begingroup$

    $a_n=(n!)^2$ using induction. Try to evaluate the first few terms on your own.






    share|cite|improve this answer









    $endgroup$



    $a_n=(n!)^2$ using induction. Try to evaluate the first few terms on your own.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 26 at 23:30









    abc...abc...

    3,237738




    3,237738












    • $begingroup$
      This is the only reasonable answer, but it shows that the problem is somewhat stupid as $n!$ is defined literally like that, i.e. $n! = n cdot (n-1) cdot ldots cdot 2 cdot 1$...
      $endgroup$
      – Klaus
      Jan 26 at 23:37




















    • $begingroup$
      This is the only reasonable answer, but it shows that the problem is somewhat stupid as $n!$ is defined literally like that, i.e. $n! = n cdot (n-1) cdot ldots cdot 2 cdot 1$...
      $endgroup$
      – Klaus
      Jan 26 at 23:37


















    $begingroup$
    This is the only reasonable answer, but it shows that the problem is somewhat stupid as $n!$ is defined literally like that, i.e. $n! = n cdot (n-1) cdot ldots cdot 2 cdot 1$...
    $endgroup$
    – Klaus
    Jan 26 at 23:37






    $begingroup$
    This is the only reasonable answer, but it shows that the problem is somewhat stupid as $n!$ is defined literally like that, i.e. $n! = n cdot (n-1) cdot ldots cdot 2 cdot 1$...
    $endgroup$
    – Klaus
    Jan 26 at 23:37













    1












    $begingroup$

    Multiply these lines together:
    begin{align}
    a_n&=n^2times a_{n-1}\
    a_{n-1}&=(n-1)^2times a_{n-2}\
    &dots\
    a_2&=2^2times a_{1}\
    a_1&=1^2times a_{0}\
    end{align}
    After canceling, we get
    $$
    a_n = 1^2 cdot 2^2 cdots n^2 a_0 = (1 cdot 2 cdots n)^2 = (n!)^2
    $$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Multiply these lines together:
      begin{align}
      a_n&=n^2times a_{n-1}\
      a_{n-1}&=(n-1)^2times a_{n-2}\
      &dots\
      a_2&=2^2times a_{1}\
      a_1&=1^2times a_{0}\
      end{align}
      After canceling, we get
      $$
      a_n = 1^2 cdot 2^2 cdots n^2 a_0 = (1 cdot 2 cdots n)^2 = (n!)^2
      $$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Multiply these lines together:
        begin{align}
        a_n&=n^2times a_{n-1}\
        a_{n-1}&=(n-1)^2times a_{n-2}\
        &dots\
        a_2&=2^2times a_{1}\
        a_1&=1^2times a_{0}\
        end{align}
        After canceling, we get
        $$
        a_n = 1^2 cdot 2^2 cdots n^2 a_0 = (1 cdot 2 cdots n)^2 = (n!)^2
        $$






        share|cite|improve this answer









        $endgroup$



        Multiply these lines together:
        begin{align}
        a_n&=n^2times a_{n-1}\
        a_{n-1}&=(n-1)^2times a_{n-2}\
        &dots\
        a_2&=2^2times a_{1}\
        a_1&=1^2times a_{0}\
        end{align}
        After canceling, we get
        $$
        a_n = 1^2 cdot 2^2 cdots n^2 a_0 = (1 cdot 2 cdots n)^2 = (n!)^2
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 27 at 0:37









        lhflhf

        166k11172402




        166k11172402






























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