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Suppose that for $1 \leq k < N$, $p_k = \dfrac{1}{2}(p_{k+1} + p_{k-1})$.

Let $b_k = p_k - p_{k-1}$. I am trying to prove that $b_k = b_{k-1}$. I've tried rearranging the terms in $p_k - p_{k-1}$ but was not able to get anywhere.

I would appreciate a small hint to point me in the right direction, thanks!

TheProofIsTrivium
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1 Answers1

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$$p_k = \frac12 (p_{k+1} + p_{k-1})$$

$$\frac12p_k + \frac12 p_k = \frac12 (p_{k+1} + p_{k-1})$$

Get rid of $\frac12$ and try to express things in terms of $b$.

Siong Thye Goh
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  • Got it, thank you. $\frac{1}{2}(p_k + p_k) = p_k = \frac{1}{2}(p_{k+1} + p_{k-1}) \Rightarrow p_k - p_{k-1} = p_{k+1} - p_k$. That was embarrassing. – TheProofIsTrivium Sep 13 '18 at 05:54