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The question ask us to guess an explicit formula for the sequence

$$s_k = s_{k-1} + 2k ,$$ for all integers $k$ greater than or equal to one and $s_0 = 3$

Can someone help me with this? Because I don't really understand how to do this. Any help will be appreciated.

Mark Fantini
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    The first step in guessing a formula is to collect numerical data! Can you at least work out the first 10 terms of the sequence yourself? If not, then you don't understand what a recursively defined sequence is and you should reread your textbook or course notes. – KCd Jul 06 '14 at 04:40
  • But from my course notes, they look something like this: sk = s(k-1) + 2k; s(k-1) = s(k-2) + 2(k-1) and so on – user3676752 Jul 06 '14 at 04:45
  • You should start finding some terms to see what's going on. – Mhenni Benghorbal Jul 06 '14 at 04:45
  • @user3676752: I mean find actual numbers. By definition $s_0$ is $3$. And what is $s_1$? What is $s_2$? What is $s_3$? Use the recursion to figure out the first few terms of the sequence as real, specific, numbers. Writing $s_{k-1} = s_{k-2} + 2(k-1)$ is not progress for collecting numerical data. – KCd Jul 06 '14 at 04:48
  • Otherwise here is a technique you can use. – Mhenni Benghorbal Jul 06 '14 at 04:52

2 Answers2

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Hint: Moving the $s_{k-1}$ term to the LHS of the equation, the recurrence relation reads:

$$s_{k}-s_{k-1}=2k.$$

You can solve this recurrence relation by simply summing both sides over $k$:

$$\sum_{k=1}^{\infty}(s_{k}-s_{k-1})=2\sum_{k=1}^{\infty}k.$$

The sum on the LHS telescopes, while the sum on the RHS is a straightforward arithmetic progression. There are standard methods for finding nice closed forms for both series.

David H
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Rearrange to get:

$$s_k - s_{k - 1} = 2k$$

If we do a summation on both sides, a humongous amount of cancellation occurs on the LHS:

$$\sum_{i = 1}^{k} (s_i - s_{i - 1}) = \sum_{i = 1}^k 2i$$ $$s_k - s_0 = 2\sum_{i = 1}^ki$$

The RHS is easy to evaluate. Now just shift $s_0$ over to get $s_k$.

Yiyuan Lee
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