rekurencja

[parchimus]

Rozwiąż rekurencje.
\(\displaystyle{ a_{0} = 1, a_{1} = 2\\
a_{n}=-a_{n-2}+1}\)

[kerajs]

\(\displaystyle{ \begin{cases} a_{4n}=1 \\ a_{4n+1}=2 \\ a_{4n+2}=0 \\ a_{4n+3}=-1 \end{cases} }\)

[Mariusz M]

\(\displaystyle{ a_{0} = 1, a_{1} = 2\\
a_{n}=-a_{n-2}+1\\
A\left( x\right) = \sum_{n=0}^{ \infty }a_{n}x^{n} \\
\sum_{n=2}^{ \infty }a_{n}x^{n} = \sum_{n=2}^{ \infty }\left( -a_{n-2}\right) x^{n} + \sum_{n=2}^{ \infty }x^{n} \\
\sum_{n=2}^{ \infty }a_{n}x^{n} =-x^{2}\left(\sum_{n=2}^{ \infty }a_{n-2}x^{n-2} \right) + \frac{x^2}{1-x} \\
\sum_{n=0}^{ \infty }a_{n}x^{n}-1-2x =-x^{2}\left(\sum_{n=0}^{ \infty }a_{n}x^{n} \right) + \frac{x^2}{1-x} \\
\left( 1+x^2\right)A\left( x\right)=1+2x+ \frac{x^2}{1-x} \\
\left( 1+x^2\right)A\left( x\right)=\frac{\left( 1+2x\right)\left( 1-x\right)+x^2 }{\left( 1-x\right) }\\
A\left( x\right) = \frac{1-x+2x-2x^2+x^2}{\left( 1-x\right)\left( 1+x^2\right) } \\
A\left( x\right) = \frac{1+x-x^2}{\left( 1-x\right)\left( 1+x^2\right) } \\
\frac{1+x-x^2}{\left( 1-x\right)\left( 1+x^2\right) } =\frac{A}{1-x}+ \frac{B}{1-ix} + \frac{C}{1+ix} \\
1+x-x^2=A\left( 1+x^2\right) +B\left( 1-x\right)\left( 1+ix\right) +C\left( 1-x\right)\left( 1-ix\right) \\

}\)

\(\displaystyle{ a_{n}= \frac{1}{2} + \frac{ \sqrt{10} }{2}\cos{\left( \frac{\pi}{2}n-\arctan\left( 3\right) \right) } }\)