Respuesta :
the first expression :
[tex]\begin{gathered} f(n)=f(n-1)+2;\text{ n}\ge2, \\ \text{where we have f(1)=}8 \end{gathered}[/tex]so at n=2,
[tex]\begin{gathered} f(2)=f(2-1)+2 \\ f(2)=f(1)+2 \\ f(2)=8+2_{} \\ f(2)=10 \end{gathered}[/tex]put n=2 in the equation f(n) = 2n +6
[tex]\begin{gathered} f(n)=2n+6 \\ f(2)=4+6 \\ f(2)=10 \end{gathered}[/tex]so,
f(n)=f(n-1)+2 is equal to f(n) =2n +6
Now for expression 2 :
[tex]\begin{gathered} f(n)=f(n-1)-2;\text{ n}\ge2 \\ \text{where }f(1)=8 \\ \text{put n=2 in the given expression,} \\ f(2)=f(2-1)-2 \\ f(2)=f(1)-2 \\ f(2)=8-2 \\ f(2)=6 \end{gathered}[/tex]put n=2 in the expression f(n)=-2n+10
[tex]\begin{gathered} f(n)=-2n+10 \\ f(2)=-2\times2+10 \\ f(2)=-4+10 \\ f(2)=6 \end{gathered}[/tex]In both the expression at n=2, f(2)=6 so,
f(n)=f(n-1)-2 is equal to f(n)=-2n+10.
Now expression 3:
[tex]\begin{gathered} f(n)=f(n-1)-3;\text{ n}\ge2 \\ \text{where , f(}1)=9 \end{gathered}[/tex]put n=2,
[tex]\begin{gathered} f(2)=f(2-1)-3 \\ f(2)=f(1)-3 \\ f(2)=9-3 \\ f(2)=6 \end{gathered}[/tex]Now for the expression at the entry of third coloumn,
f(n)=-3n+12
put n=2
[tex]\begin{gathered} f(n)=-3n+12 \\ f(2)=-3(2)+12 \\ f(2)=-6+12 \\ f(2)=6 \end{gathered}[/tex]since, the f(2) =6,
