Fibonacci sequence squared

Task number: 3318

Prove that for every integer \(n\ge 4\) \(\displaystyle F_n^2=2F_{n-1}^2+2F_{n-2}^2-F_{n-3}^2.\)

  • Hint

    Express \(F_{n-3}\) using \(F_{n-1}\) and \(F_{n-2}\).

  • Solution

    \[\begin{align*} 2F_{n-1}^2+2F_{n-2}^2-F_{n-3}^2 &= 2F_{n-1}^2+2F_{n-2}^2-(F_{n-1}-F_{n-2})^2 \\ &= 2F_{n-1}^2+2F_{n-2}^2-(F_{n-1}^2-2F_{n-2}F_{n-1}+F_{n-2}^2) \\ &= F_{n-1}^2+F_{n-2}^2+2F_{n-2}F_{n-1}\\ &= (F_{n-1}+F_{n-2})^2\\ &= F_n^2 \end{align*}\]

    Notice that we never used the values of \(F_1\) and \(F_2\). So the given equation holds for all recurrent sequences satisfying \(F_n=F_{n-1}+F_{n-2}\).

Difficulty level: Moderate task
Solution require uncommon idea
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