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For a sequence $\{D_k\}$, if we have: $$D_k=pD_{k+1}+qD_{k-1}+1$$

and we know that $D_0=D_N=0$. Where $p+q=1$, and $N$ is known. How do I solve it?

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closed as off-topic by choco_addicted, kamil09875, hardmath, Shailesh, Jonas Mar 12 at 0:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – choco_addicted, kamil09875, hardmath, Shailesh, Jonas
If this question can be reworded to fit the rules in the help center, please edit the question.

Here is a related problem. – Mhenni Benghorbal Nov 4 '12 at 19:02
I would use a generating function: Let $d(x) = \sum x^k D_k$, apply the recurrence and end conditions, and see what happens. – marty cohen Nov 4 '12 at 20:17

First assume $p \neq q$. Define $E_k = D_k + \frac{k}{p - q}$. Then $E$ satisfies the linear recursion

$$ E_k = p E_{k+1} + q E_{k - 1}. $$

A solution of this recursion is of the form $E_k = \alpha \left(\frac{q}{p}\right)^k + \beta$ for constants $\alpha, \beta$. Using that $D_0 = D_N = 0$ this results in

$$ D_k = \frac{N}{p - q} \cdot \frac{\left(\frac{q}{p}\right)^k - 1}{\left(\frac{q}{p}\right)^N - 1} - \frac{k}{p-q}. $$

If $p = q = \tfrac{1}{2}$ then define $E_k = D_k + k^2$ which satisfies the recursion

$$ E_k = \tfrac{1}{2} E_{k+1} + \tfrac{1}{2} E_{k-1}. $$

Now a solution is of the form $E_k =\alpha \, k + \beta$. In this case we find

$$ D_k = k \, (N - k). $$

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