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finding a solution for this recurrence relation

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finding a solution for this recurrence relation


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4












$begingroup$



Find the sequence satisfying the recurrence relation $$n a_{n+1} = (n+1) a_n+n(n+1)$$ with the initial condition $a_0=0$.




I'm trying to find a solution for this recurrence relation. After dividing both sides by $n(n+1)$, we get:
$$frac{a_n+1}{n+1} =frac{a_n}{n} +1.$$



After setting $b_k=a_n/n$, the resulting relation is $b_{k+1} = b_k +1$.



$b_{k+2} = b_{k+1} +1$
by subtracting this relation from the one before it, I got :



$b_{k+2} -2*b_{k+1} +b_k = 0$
after that I replace $b_k = lambda^k$
I got :



$lambda^{k+2} -2*lambda^{k+1}+b_k$ and after dividing the relation by $lambda^k$ the result is $lambda^2 + lambda +1 = 0$



and the solution for the last relation is $lambda = 1$



but I stopped there because the initial Condition is
$a_0=0$ and I could't find the value of $a_1$.



can someone show me a proper way or a proper solution for this recurrence relation?






recurrence-relations recursion initial-value-problems







share|cite|improve this question









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Ahmad Rmmo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$












  • $begingroup$
    $b_{k+1} = b_k +1$ gives $b_k = k -1 + b_1$.
    $endgroup$
    – lhf
    35 mins ago
















4












$begingroup$



Find the sequence satisfying the recurrence relation $$n a_{n+1} = (n+1) a_n+n(n+1)$$ with the initial condition $a_0=0$.




I'm trying to find a solution for this recurrence relation. After dividing both sides by $n(n+1)$, we get:
$$frac{a_n+1}{n+1} =frac{a_n}{n} +1.$$



After setting $b_k=a_n/n$, the resulting relation is $b_{k+1} = b_k +1$.



$b_{k+2} = b_{k+1} +1$
by subtracting this relation from the one before it, I got :



$b_{k+2} -2*b_{k+1} +b_k = 0$
after that I replace $b_k = lambda^k$
I got :



$lambda^{k+2} -2*lambda^{k+1}+b_k$ and after dividing the relation by $lambda^k$ the result is $lambda^2 + lambda +1 = 0$



and the solution for the last relation is $lambda = 1$



but I stopped there because the initial Condition is
$a_0=0$ and I could't find the value of $a_1$.



can someone show me a proper way or a proper solution for this recurrence relation?






recurrence-relations recursion initial-value-problems







share|cite|improve this question









New contributor




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







$endgroup$












  • $begingroup$
    $b_{k+1} = b_k +1$ gives $b_k = k -1 + b_1$.
    $endgroup$
    – lhf
    35 mins ago














4












4








4


2



$begingroup$



Find the sequence satisfying the recurrence relation $$n a_{n+1} = (n+1) a_n+n(n+1)$$ with the initial condition $a_0=0$.




I'm trying to find a solution for this recurrence relation. After dividing both sides by $n(n+1)$, we get:
$$frac{a_n+1}{n+1} =frac{a_n}{n} +1.$$



After setting $b_k=a_n/n$, the resulting relation is $b_{k+1} = b_k +1$.



$b_{k+2} = b_{k+1} +1$
by subtracting this relation from the one before it, I got :



$b_{k+2} -2*b_{k+1} +b_k = 0$
after that I replace $b_k = lambda^k$
I got :



$lambda^{k+2} -2*lambda^{k+1}+b_k$ and after dividing the relation by $lambda^k$ the result is $lambda^2 + lambda +1 = 0$



and the solution for the last relation is $lambda = 1$



but I stopped there because the initial Condition is
$a_0=0$ and I could't find the value of $a_1$.



can someone show me a proper way or a proper solution for this recurrence relation?






recurrence-relations recursion initial-value-problems







share|cite|improve this question









New contributor




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







$endgroup$





Find the sequence satisfying the recurrence relation $$n a_{n+1} = (n+1) a_n+n(n+1)$$ with the initial condition $a_0=0$.




I'm trying to find a solution for this recurrence relation. After dividing both sides by $n(n+1)$, we get:
$$frac{a_n+1}{n+1} =frac{a_n}{n} +1.$$



After setting $b_k=a_n/n$, the resulting relation is $b_{k+1} = b_k +1$.



$b_{k+2} = b_{k+1} +1$
by subtracting this relation from the one before it, I got :



$b_{k+2} -2*b_{k+1} +b_k = 0$
after that I replace $b_k = lambda^k$
I got :



$lambda^{k+2} -2*lambda^{k+1}+b_k$ and after dividing the relation by $lambda^k$ the result is $lambda^2 + lambda +1 = 0$



and the solution for the last relation is $lambda = 1$



but I stopped there because the initial Condition is
$a_0=0$ and I could't find the value of $a_1$.



can someone show me a proper way or a proper solution for this recurrence relation?






recurrence-relations recursion initial-value-problems




recurrence-relations recursion initial-value-problems






share|cite|improve this question









New contributor




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











share|cite|improve this question









New contributor




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









share|cite|improve this question




share|cite|improve this question








edited 4 hours ago









Abdalrohman Alsalkhadi

52




52






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asked 5 hours ago









Ahmad RmmoAhmad Rmmo

212




212




New contributor




Ahmad Rmmo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





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






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












  • $begingroup$
    $b_{k+1} = b_k +1$ gives $b_k = k -1 + b_1$.
    $endgroup$
    – lhf
    35 mins ago


















  • $begingroup$
    $b_{k+1} = b_k +1$ gives $b_k = k -1 + b_1$.
    $endgroup$
    – lhf
    35 mins ago
















$begingroup$
$b_{k+1} = b_k +1$ gives $b_k = k -1 + b_1$.
$endgroup$
– lhf
35 mins ago




$begingroup$
$b_{k+1} = b_k +1$ gives $b_k = k -1 + b_1$.
$endgroup$
– lhf
35 mins ago










2 Answers
2






active

oldest

votes


















5












$begingroup$

This is not a well-posed problem. In other words, given $a_0=0$, the value of $a_1$ can be any real number whatsoever since it must satisfy
$$
0 cdot a_1 = 1 cdot a_0 + 0 cdot 1,
$$
and both sides are $0$ for any value of $a_1$, so the equality holds. In other words, the specified conditions do not uniquely specify a sequence.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Given the value of $a_1$ you have $a_n = n(n-1) + na_1$ for all $n$. The recursion equation
    $$n a_{n+1} = (n+1) a_n+n(n+1)tag{1}$$ for $n=0$ is just $a_0=0.$ In other words, you should start with initial condition for $a_1$ and then all other values of $a_n$ for $n>1$ are determined, and also $a_0=0.$






    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









      5












      $begingroup$

      This is not a well-posed problem. In other words, given $a_0=0$, the value of $a_1$ can be any real number whatsoever since it must satisfy
      $$
      0 cdot a_1 = 1 cdot a_0 + 0 cdot 1,
      $$
      and both sides are $0$ for any value of $a_1$, so the equality holds. In other words, the specified conditions do not uniquely specify a sequence.






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        This is not a well-posed problem. In other words, given $a_0=0$, the value of $a_1$ can be any real number whatsoever since it must satisfy
        $$
        0 cdot a_1 = 1 cdot a_0 + 0 cdot 1,
        $$
        and both sides are $0$ for any value of $a_1$, so the equality holds. In other words, the specified conditions do not uniquely specify a sequence.






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          This is not a well-posed problem. In other words, given $a_0=0$, the value of $a_1$ can be any real number whatsoever since it must satisfy
          $$
          0 cdot a_1 = 1 cdot a_0 + 0 cdot 1,
          $$
          and both sides are $0$ for any value of $a_1$, so the equality holds. In other words, the specified conditions do not uniquely specify a sequence.






          share|cite|improve this answer









          $endgroup$



          This is not a well-posed problem. In other words, given $a_0=0$, the value of $a_1$ can be any real number whatsoever since it must satisfy
          $$
          0 cdot a_1 = 1 cdot a_0 + 0 cdot 1,
          $$
          and both sides are $0$ for any value of $a_1$, so the equality holds. In other words, the specified conditions do not uniquely specify a sequence.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 5 hours ago









          gt6989bgt6989b

          36.3k22557




          36.3k22557























              0












              $begingroup$

              Given the value of $a_1$ you have $a_n = n(n-1) + na_1$ for all $n$. The recursion equation
              $$n a_{n+1} = (n+1) a_n+n(n+1)tag{1}$$ for $n=0$ is just $a_0=0.$ In other words, you should start with initial condition for $a_1$ and then all other values of $a_n$ for $n>1$ are determined, and also $a_0=0.$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Given the value of $a_1$ you have $a_n = n(n-1) + na_1$ for all $n$. The recursion equation
                $$n a_{n+1} = (n+1) a_n+n(n+1)tag{1}$$ for $n=0$ is just $a_0=0.$ In other words, you should start with initial condition for $a_1$ and then all other values of $a_n$ for $n>1$ are determined, and also $a_0=0.$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Given the value of $a_1$ you have $a_n = n(n-1) + na_1$ for all $n$. The recursion equation
                  $$n a_{n+1} = (n+1) a_n+n(n+1)tag{1}$$ for $n=0$ is just $a_0=0.$ In other words, you should start with initial condition for $a_1$ and then all other values of $a_n$ for $n>1$ are determined, and also $a_0=0.$






                  share|cite|improve this answer









                  $endgroup$



                  Given the value of $a_1$ you have $a_n = n(n-1) + na_1$ for all $n$. The recursion equation
                  $$n a_{n+1} = (n+1) a_n+n(n+1)tag{1}$$ for $n=0$ is just $a_0=0.$ In other words, you should start with initial condition for $a_1$ and then all other values of $a_n$ for $n>1$ are determined, and also $a_0=0.$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 51 mins ago









                  SomosSomos

                  15.6k11437




                  15.6k11437






















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