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?
On
The second column of the table will be
$$-1,3,1,f (4)$$ the third is
$$4,-2,f (4)-1$$
the fourth gives
$$(f (4)-1)-(-2)=6 \implies f (4)=5$$
On
The key is to observe that each cell in the table is the sum of the value immediately above it and the value above and to the right. So for example, $$f(2) = f(1) + \Delta f(1) = -1 + 4 = 3.$$ We continue to compute: $$f(3) = 3 - 2 = 1, \\ \Delta f(3) = -2 + 6 = 4, \\ f(4) = 1 + 4 = 5,$$ and we are done. Note it is not necessary to compute $\Delta^2 f(1)$.
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 $f(4) = 5$.