Aby wyznaczyć n+1-szy wyraz należy w miejsce n wstawić n+1.

a) Mamy

$a_{n+1}=2{\left(n+1\right)}^2+4\left(n+1\right)+2=2\left(n^2+2n+1\right)+4n+4+2=2n^2+4n+2+4n+6=2n^2+8n+8$  

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b) Mamy

$a_{n+1}=\frac{3\left(n+1\right)+5}{n+1-1}=\frac{3n+3+5}{n}=\frac{3n+8}{n}$  

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c) Mamy

$a_{n+1}=\frac{\left(n+1-3\right)\left(n+1+3\right)}{3\left(n+1\right)}=\frac{\left(n-2\right)\left(n+4\right)}{3n+3}$  

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d) Mamy

$a_{n+1}=\frac{{\left(-5\right)}^{n+1}}{-5\left(n+1\right)}=\frac{\cancel{\left(-5\right)}\cdot {\left(-5\right)}^n}{\cancel{\left(-5\right)}\cdot \left(n+1\right)}=\frac{{\left(-5\right)}^n}{n+1}$