<s>[INST] How can I find a polynomial that fits a given table? 
The first difference is 1, 4, 7, 10, 13, 16
The second difference is 3, 3, 3, 3, 3, 3
Since the second difference is constant this would be quadratic and I would have 
$\frac{3}{2}n^{2}$
So now I will take the differences between the original sequence and the values of $\frac{3}{2}n^{2}$ for n = 1, 2, ... 
3 - 3/2, 4 - 6, 8 - 27/2, 15 - 24, 25 - 75/2 
Which is 3/2, -2, -11/2, -9, -25/2
so the difference between each of those terms are 
$-3 \frac{1}{2}, -3 \frac{1}{2}, -3 \frac{1}{2}, -3 \frac{1}{2}, -3 \frac{1}{2}$
So then I have $\frac{3}{2} n^{2} -3.5n+5$ but when I plug in zero i get 5, and I should get 3, so what did I do wrong?
 [/INST] Let $\Delta_k$ be the $k^{\text{th}}$ finite diffetence. then 
$$P(x)=P(x_0)+$$
$$\sum_{k=1}^n\frac{\Delta_k}{k!}(x-x_0)(x-x_1)...(x-x_{k-1})=$$
$$3+1.(x-x_0)+\frac 32(x-x_0)(x-x_1)=$$
$$3+(x-0)+\frac 32(x-0)(x-1)=$$
$$3+x+\frac 32x^2-\frac 32x=$$
$$\frac 32x^2-\frac 12x+3.$$
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