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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 :

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

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?

SlaMath
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Ahmad Rmmo
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    $b_{k+1} = b_k +1$ gives $b_k = k -1 + b_1$. – lhf May 03 '19 at 00:33
  • Though of course you should be kicking yourself for missing the obvious solution lhf has pointed out, what else you missed is that 2nd order recursions have 2 independent solutions, not just 1. And the one you found is the false solution you introduced by converting it from a 1st order recursion to a 2nd order recursion, since $b_n = 1$ obviously does not satisfy $b_{k+1} = b_k + 1$. – Paul Sinclair May 03 '19 at 04:05

2 Answers2

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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.

gt6989b
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Given the value of $a_1$ you have $a_n = n(n-1) + na_1$ for all $n\ge 0.$ 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 $a_0=0.$

If you want to solve for $n<0$, then given the value of $a_{-1}$ you have the equation $a_n = n(n-1) + na_{-1}$ which determines $a_n$ for all $n<0.$

Somos
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