(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let [tex]\frac{1}{1-z+z^2} = \sum_{n=0}^{\infty} a_n \cdot z^n[/tex]

where [tex]a_{1} = a_0 = 1 \Leftrightarrow a_{1}-a_{0} = 0[/tex] and [tex]a_2=0[/tex] and [tex]a_{n+3} + a_n = 0[/tex]

3. The attempt at a solution

By the theorem of Identity

[tex]1 = 1 - z + z^2 \sum a_n \cdot z^n = a_0 + (a_{1}-a_{0})z + a_{2} \cdot z^2 + \sum_{n=3}^{\infty} a_n \cdot z^{n} - \sum_{n=1}^{\infty} a_{n} \cdot z^{n+1} + \sum_{n=0}^{\infty} a_n z^{n+2}[/tex]

If I write out the first sum I get [tex]\sum_{n=3}^{\infty} a_n \cdot z^{n} = a_3 z^3 + a_{4}z^4 + \cdots [/tex]

I write out a couple terms of the second sum [tex]- \sum_{n=1}^{\infty} a_{n} \cdot z^{n+1} = - a_1 \cdot z^{2} - a_2 \cdot z^{3} - \cdots [/tex]

If I make the commen index n = 0

it looks like the index increase at n+3 and at a power of n+2 such that

[tex]a_0 + (a_{1}-a_{0})z + a_{2} \cdot z^2 +\sum_{n=3}^{\infty} + a_n \cdot z^{n} - \sum_{n=1}^{\infty} a_{n} \cdot z^{n+1} + \sum_{n=0}^{\infty} a_n z^{n+2} = \sum_{n=0}^{\infty} (a_{n+3} +a_n)z^{n+2}[/tex]

How is this?

Sincerely

Susanne

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