Wykaż, że ciąg jest arytmetyczny.
a_{n} = \frac {2n^2 + n - 6} {n+2} \\ a_{n+1}-a_{n} = \frac {(2n+1)^2 + n +1 - 6} {n +1+2} - \frac {2n^2 + n - 6} {n+2} \\ = \frac{4n^2 + 1+n-5}{ n+3}- \frac {2n^2 + n - 6} {n+2} \\ = \frac{(4n^2+n-4)(n+2)}{(n+3)(n+2)}-\frac{(2n^2 + n - 6)(n+3)}{(n+3)(n+2)} \\ = \frac{4n^3 +8n^2+n^2 + 2n -4n - 8}{(n+3)(n+2)} - \frac{2n^3+6n^2+n^2 +3n - 6n - 18}{(n+3)(n+2)} \\ = \frac{2n^3+ 2n^2+n +10}{(n+3)(n+2)}
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a_{n} = \frac {2n^2 + n - 6} {n+2} \\ a_{n+1}-a_{n} = \frac {(2n+1)^2 + n +1 - 6} {n +1+2} - \frac {2n^2 + n - 6} {n+2} \\ = \frac{4n^2 + 1+n-5}{ n+3}- \frac {2n^2 + n - 6} {n+2} \\ = \frac{(4n^2+n-4)(n+2)}{(n+3)(n+2)}-\frac{(2n^2 + n - 6)(n+3)}{(n+3)(n+2)} \\ = \frac{4n^3 +8n^2+n^2 + 2n -4n - 8}{(n+3)(n+2)} - \frac{2n^3+6n^2+n^2 +3n - 6n - 18}{(n+3)(n+2)} \\ = \frac{2n^3+ 2n^2+n +10}{(n+3)(n+2)}