Odpowiedź:
[tex]r=-\frac{1}{2}[/tex]
Szczegółowe wyjaśnienie:
[tex]a_n=\frac{1}{2}(4-n)=2-\frac{1}{2}n\\\\a_{n+1}=2-\frac{1}{2}(n+1)=2-\frac{1}{2}n-\frac{1}{2}=1\frac{1}{2}-\frac{1}{2}n\\\\r=a_{n+1}-a_n=\left(1\frac{1}{2}-\frac{1}{2}n\right)-\left(2-\frac{1}{2}n\right)=1\frac{1}{2}-\frac{1}{2}n-2+\frac{1}{2}n=-\frac{1}{2}[/tex]
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Odpowiedź:
[tex]r=-\frac{1}{2}[/tex]
Szczegółowe wyjaśnienie:
[tex]a_n=\frac{1}{2}(4-n)=2-\frac{1}{2}n\\\\a_{n+1}=2-\frac{1}{2}(n+1)=2-\frac{1}{2}n-\frac{1}{2}=1\frac{1}{2}-\frac{1}{2}n\\\\r=a_{n+1}-a_n=\left(1\frac{1}{2}-\frac{1}{2}n\right)-\left(2-\frac{1}{2}n\right)=1\frac{1}{2}-\frac{1}{2}n-2+\frac{1}{2}n=-\frac{1}{2}[/tex]