We can test each of the options in order to see if we can discard the ones that doesn't represent the table shown in the relation.
We will test the first option for n=x=2:
[tex]\begin{gathered} f(n)=f(n-1)-2f(1) \\ n=2 \\ f(2)=f(1)-2f(1)=-2-2(-2)=-2+4=2\neq-2 \end{gathered}[/tex]
That doesn't match the table result, so we can discard this option.
We test the second option the same way:
[tex]\begin{gathered} f(n)=f(n+1)-2f(1) \\ n=2 \\ f(2)=f(3)-2f(1)=-6-2(-2)=-6+4=-2 \end{gathered}[/tex]
This match the result from the table, so it can be the answer. But we have to test the other functions to see if we can discard them:
The third option is:
[tex]\begin{gathered} f(n)=f(n-1)+2f(1) \\ n=2 \\ f(2)=f(1)+2f(1)=-2+2(-2)=-2-4=-6\neq-2 \end{gathered}[/tex]
Discarded.
The fourth option is:
[tex]\begin{gathered} f(n)=f(n-1)\cdot2f(1) \\ n=2 \\ f(2)=f(1)\cdot2f(1)=-2\cdot2(-2)=-2\cdot(-4)=8\neq-2 \end{gathered}[/tex]
Discarded too.
The only function that was not discarded was the second one, so it has to be the answer.
This is because at least one of the options has to be correct. If not, we should have test for each of the table values in order to be sure.
