what does the common difference of a sequence describe I am given the sequence 1, 7, 21, 43, 73, ... and can derive without any assumption about the nature of the sequence that the ultimate difference is 8, I'm interested in understanding what this difference allows you to deduce about the original polynomial which generates this sequence; and whether it is possible to find this polynomial?
I can see that 1, 7, 21, 43, 73, ... has a difference of 6, 14, 22, 30, ... which is generated by the arithmetic sequence 6 + 8(n-1), but trying to rationalize how the difference of a sequence can then be non-constant (1, 7, 21, 43, 73) it doesn't really make sense.
Can anyone shed some light on how to go about solving this intuitively? 
 A: In order to find the polynomial that generated the sequence, you need to look at differences between individual terms inside the sequence. You are given that the first few terms of the sequence are
$$(1, 7, 21, 43, 73, \dots)$$
Therefore, you need to start computing differences between terms. First, observe that the difference between the initial terms isn't constant. Although, if we take the second difference between these terms then we form the constant value of $8$.
$$\text{Second Difference:} ~~~~~~~~~8 ~~~~ 8 ~~~~8 \dots~~~~~~$$
$$\text{First Difference:}~~~~~~~~6 ~~ 14 ~~ 22 ~~ 30 \dots$$
$$\text{Original Sequence:}~~~~1 ~~7 ~~21 ~~43 ~~73 \dots$$
As the second difference is constant, we have a quadratic sequence. The general form of a quadratic is 
$$y = ax^2 + bx+c$$
The first five terms produce
$$1 = a + b + c\tag{1}$$
$$7 = 4a + 2b + c\tag{2}$$
$$21= 9a + 3b + c\tag{3}$$
$$43= 16a + 4b + c\tag{4}$$
$$73= 25a + 5b + c\tag{5}$$
Taking $(1)-(2)$ forms
$$-6 = -3a -b\tag{6}$$
and $(3)-(2)$ forms
$$14 = 5a + b\tag{7}$$
so that $(6)+(7)$ forms $a=4$. By $(7)$, we see that $14 = 20 + b$ and so $b=-6$. Another substitution back into $(1)$ will form $c=3$. Hence, the equation of the polynomial is
$$y = 4x^2 - 6x + 3$$
In general, you shouldn't try to find the generating polynomial unless you have more terms and can see a consistent pattern between the differences between terms.
A: The difference sequence is $6, 14, 22, 30,$ etc.,  for which, as you noted, the $n^{th}$ term is $6+8(n-1)$.  
When you evaluate the sum for the $m^{th}$ term of the sequence, $1+\sum\limits_{n=1}^{m-1}[6+8(n-1)], $
the answer is $1+8\sum\limits_{n=1}^{m-1}n-\sum\limits_{n=1}^{m-1}2=1+8 \dfrac{(m-1)m}2-2(m-1)=4m^2-6m+3. $
Just as when you integrate a $k^{th}$ degree polynomial, the answer is a polynomial of degree $k+1$, 
when you repeatedly sum a $k^{th}$ degree (in this case, linear) polynomial in $n$ from $1$ to $m-1$, 
the answer is a ${k+1}^{th}$ degree (in this case, quadratic) polynomial in $m$.
A: You have to go through $2$ levels of differences to get to the ultimate difference, so your polynomial will be of degree $2$. (If you know some calculus, this corresponds to taking $2$ derivatives to reach a constant function.)
Because the ultimate difference is $8$ and it appears at degree $2$, the first coefficient is $\frac8{2!}=4$ (the $2!$ is again a factor that appears if you differentiate; if you don't know any calculus, I'm not sure if there's an intuitive explanation for it, but you can derive it by a method similar to Axion004's answer), so the first term of the polynomial is $4x^2$.
The first term of the polynomial will get in the way of determining the other terms intuitively, so at this point it's easiest to just subtract it out. Subtracting $4x^2$ from each term (and assuming the terms are indexed starting at $x=0$) gives
$$1,3,5,7,9,\dots$$
Now we go through the same process again. We know the degree will now go down one level to $1$, and we can find that the new ultimate difference is $2$, so the next coefficient is $\frac2{1!}=2$, and so far our polynomial is $4x^2+2x$. Subtracting the $2x$ from our last sequence gives 
$$1,1,1,1,1,\dots$$
so we have reached the constant sequence. Our generating polynomial is $4x^2+2x+1$.
