Letting \(P_{n}\) be the \(n\)th Pell number:

\[{P_{n}}^2=\frac{(-1)^{n+1}}{4}+\sum_{k=0}^{\frac{n-(n\bmod2)}{2}}{n\choose2k}\cdot\Bigg(\frac{3^n\cdot8^k}{4\cdot9^k}\Bigg)\]\[{P_{n}}^4=\frac{3}{32}+\sum_{k=0}^{\frac{n-(n\bmod2)}{2}}{n\choose2k}\cdot\Bigg(\frac{17^n\cdot288^k}{32\cdot289^k}-\frac{(-3)^n\cdot8^k}{8\cdot9^k}\Bigg)\]

I made a GIF to illustrate sequence https://oeis.org/A028401 and I felt like playing around with that patternโฆ https://mathstodon.xyz/media/IuKBtWbSKIVIlYYzrHc

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#proofinatoot that \(\sqrt{2}\) is irrational:

Suppose not, i.e. \( \sqrt{2} = \frac{a}{b} \), \( a,b \in \mathbb{Z} \) and \(a,b\) are coprime.

Then \( \left(\frac{a}{b}\right)^2 = 2\). Then \(a^2 = 2b^2\), so \(a^2\) is even. Let \(a = 2c\), so we have \(4c^2 = 2b^2 \implies b^2 = 2c^2\). So \(b\) is even.

Then both \(a\) and \(b\) are divisible by 2, contradicting the earlier proposition.

So \(\sqrt{2}\) can't be expressed as the ratio of two integers, which means it's irrational.

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๐ณ๏ธโ๐ An otter human being here. INTJ-A 5w6. Checking what everyone takes for granted. Joking around. By the way, I love maths and puns.

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