Q:

Subject math GRE exam 9768 Q.17

The question is as follows:

And the answer is E.
I calculated the following:
$\Delta^2f(1) = -6, f(2) = 3, f(3)=1, \Delta f(3)=4.$
Then I stopped and don`t know how to complete because the formula for calculating $\Delta^2f(3)$ contain $\Delta f(4)$. So what shall I do could anyone help me please? 

A:

The second column of the table will be
$$-1,3,1,f (4)$$
the third is
$$4,-2,f (4)-1$$
the fourth gives
$$(f (4)-1)-(-2)=6 \implies f (4)=5$$

A:

Based off the table: $f(1) = 1$ and $\Delta f(1) = f(2) - f(1 ) = 4$ so we have $f(2) = 3$. By $\Delta f(2) = f(3) - f(2) = -2$, we have $f(3) = 1$. Further spell out $\Delta^2 f(2) = \Delta f(3) - \Delta f(2) = f(4) - f(3) + 2 = 6$, substitute already calculated $f(3)$, you would get $f(4) = 5$.

