Solve the following recurrence relation:
$$l(n+1)=l(n)+π(4+2cn), \quad n=0,1,2,3,\ldots$$
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4Welcome to Math.SE. Can you tell us what you've tried in order to solve this? – Sujaan Kunalan Jun 06 '13 at 01:32
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i can solve linear reccurence relations but l(n+1)=l(n)+ n types equation i couldnt find enough source and i guess my problem is nonlineer one after hours of search i thought i need some help – Ahmet Atcı Jun 06 '13 at 01:42
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What does $\pi(4+2cn)$ represent in this question? The constant $\pi=3.14\ldots$ multiplied by $4+2cn$ where $c\in \mathbb{R}$ is some constant? Or perhaps $\pi(\ldots)$ is the prime counting function? Hopefully by $\pi$ you don't mean some group-theoretic object? – Douglas B. Staple Jun 06 '13 at 02:04
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i guess c is constant but there is no other information in the paper about this question so i dont know what π represents either. but i will take it as a constant too. – Ahmet Atcı Jun 06 '13 at 02:09
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See here for general techniques. – Mhenni Benghorbal Jun 10 '13 at 20:20
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You can rewrite the equation as: $$ l(n + 1) - l(n) = \pi(4 + 2 c n) $$ Thus: $$ \begin{align*} \sum_{0 \le k < n} (l(k + 1) - l(k)) &= \sum_{0 \le k < n} \pi(4 + 2 c k) \\ l(n) &= l(0) + \sum_{0 \le k < n} \pi(4 + 2 c k) \end{align*} $$ Your recurrence is linear, al right.
vonbrand
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