Today I've been playing around with maths looking for patterns and formulae for even powers of Pell numbers.

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

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

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\[{P_{n}}^6 = \frac{5 \cdot (-1)^{n + 1}}{128} + \sum_{k=0}^{\frac{n - (n \bmod 2)}{2}} {n \choose 2k} \cdot \Bigg(\frac{99^n \cdot 9800^k}{256 \cdot 9801^k} - \frac{3 \cdot (-17)^n \cdot 288^k}{128 \cdot 289^k} + \frac{3^{n+1} \cdot 5 \cdot 8^k}{256 \cdot 9^k}\Bigg) \]

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I got those formulae knowing that \((a+b)^2\) \(+\) \((a-b)^2\) \(=\) \(\sum_{k=0}^{\frac{n - (n \bmod 2)}{2}}\) \(2\) \({n \choose 2k}\) \(a^{n-2k}\) \(b^{2k}\).

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