<s>[INST] Use the method of finite differences to determine a formula Please evaluate my solution and let me know of anywhere I can improve or make clearer. Thank you
QUESTION:Use the method of finite differences to determine a formula for the general term of the sequence whose first seven terms are:
$$5, 19, 49, 101, 181, 295, 449, \dots$$
SOLUTION: 
The given sequence is 
$$S= 5, 19, 49, 101, 181, 295, 449, \dots$$
We can use the method of finite differences to find the general term. We write the difference of consecutive terms of the sequence till we can get a constant sequence. 
\begin{align*}
D_1 &= 14, 30, 52, 80, 84, 153, \dots \\
D_2 &= 16, 22, 28, 34, 40, \dots \\
D_3 &= 6, 6, 6, 6, \dots
\end{align*}
So $D_3$ is a constant sequence. This means its general term is a third degree polynomial.
The sequence general term is $X_n = An^3 + Bn^2 + Cn +D$
Now, we know that 
\begin{align*}
x_1 &= 5 \\
x_2 &= 19 \\
x_3 &= 49 \\
x_4 &= 101 
\end{align*}
Using $x_n = An^3 + Bn^2 + Cn +D$
We get,
\begin{equation} \label{eq1}
5 = A + B + C + D
\end{equation}
\begin{equation} \label{eq2}
19 = 8A + 4B + 2C + D
\end{equation}
\begin{equation} \label{eq3}
49 = 27A + 9B + 3C + D
\end{equation}
\begin{equation} \label{eq4}
101 = 64A + 16B + 4C + D
\end{equation}
Using equation (2) - equation (1), we get 
\begin{equation} \label{eq5}
14 = 7A + 3B + C
\end{equation}
Using equation (3) - equation (2), we get 
\begin{equation} \label{eq6}
30 = 19A + 5B + C
\end{equation}
Using equation (4) - equation (3), we get 
\begin{equation} \label{eq7}
52 = 37A + 7B + C
\end{equation}
Now subtracting again,
Using equation (6) - equation (5), we get 
\begin{equation} \label{eq8}
16 = 12A + 2B
\end{equation}
Using equation (7) - equation (6), we get 
\begin{equation} \label{eq9}
22 = 18A + 2B
\end{equation}
Again we subtract, 
Using equation (9) - equation (8), we get 
\begin{equation} \label{eq10}
6 = 6A \Rightarrow A = 1
\end{equation}
Now solving for $B, C,$ and $D$ we get
\begin{equation} \label{eq11}
22 - 18 = 2B \Rightarrow B = 2
\end{equation} 
\begin{equation} \label{eq12}
14 = 7 + 6 + C \Rightarrow C = 1
\end{equation} 
\begin{equation} \label{eq13}
5 = 1 + 2 + 1 + D \Rightarrow D = 1
\end{equation}
Hence our general term is
$$ n^3 + 2n^2 + n + 1 \text{ for } n \in \mathbb{N}$$
Solution page 1
Solution page 2
 [/INST] What you did looks basically correct.  Nonetheless, one small thing the question is implicitly stating, and you're implicitly using, is that the $D_3$ row of $6$'s will continue indefinitely. There are formulas (although they're somewhat convoluted) which give sequences where the first $7$ values match what you're given, but where not all of the later values of $D_3$ would be $6$.
Another issue is with with the terminology. As stated in the first sentence of Wikipedia's Finite difference page

A finite difference is a mathematical expression of the form $f(x + b) − f(x + a)$.

This does apply to what you're doing. However, note the Wikipedia page says at the start of the third paragraph that

Today, the term "finite difference" is often taken as synonymous with finite difference approximations of derivatives, especially in the context of numerical methods.

Also, Wolfram's MathWorld Finite difference page starts with

The finite difference is the discrete analog of the derivative.

I believe the term "finite difference" likely comes from the idea that you're using one specific, even if very small, finite difference, as opposed to differentiation which looks at using the limiting case of infinitely many small intervals to determine a difference to calculate the derivative function exactly.
Instead, as explained in Mathonline's Difference Sequences page, the $p$'th row you've determined is called the "$p^{\text{th}}$ order difference sequence". The entire set of these rows, as explained in Mathonline's Difference Tables of Sequences, is called a "difference table". This phrase is also used in other places, such as in the Example section of Wikipedia's "Binomial transform" page.
Regarding the solution method, there's an alternate method, as suggested by the Wikipedia article mentioned above, and explained quite well in the Jean-Claude Arbaut's answer to Baffled by resolving number list. Using this, assuming the index of the polynomial values starts at $0$, we get by using the binomial transform with the left-most values of $S$ and $D_i$ that
$$\begin{equation}\begin{aligned}
p_0(n) & = 5{n \choose 0} + 14{n \choose 1} + 16{n \choose 2} + 6{n \choose 3} \\
& = 5 + 14n + 16\left(\frac{n(n-1)}{2}\right) + 6\left(\frac{n(n-1)(n-2)}{6}\right) \\
& = 5 + 14n + 8(n^2 - n) + (n^3 - 3n^2 + 2n) \\
& = n^3 + 5n^2 + 8n + 5
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Converting this to using a starting index of $1$ instead gives
$$\begin{equation}\begin{aligned}
p_0(n) & = p_1(n-1) \\
& = (n-1)^3 + 5(n-1)^2 + 8(n-1) + 5 \\
& = (n^3 - 3n^2 + 3n - 1) + 5(n^2 - 2n + 1) + (8n - 8) + 5 \\
& = n^3 + 2n^2 + n + 1
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
This matches what you got. Alternatively, we can use the pattern to determine the previous values in the sequences. Starting from the $6$'s in $D_3$, we get that the previous value in $D_2$ would be $16 - 6 - 10$, the previous value in $D_1$ would be $14 - 10 = 4$ and the previous previous value in $S$ would be $5 - 4 = 1$. Thus, using the binomial transform with these values values gives
$$\begin{equation}\begin{aligned}
p_1(n) & = 1{n \choose 0} + 4{n \choose 1} + 10{n \choose 2} + 6{n \choose 3} \\
& = 1 + 4n + 10\left(\frac{n(n-1)}{2}\right) + 6\left(\frac{n(n-1)(n-2)}{6}\right) \\
& = 1 + 4n + 5(n^2 - n) + (n^3 - 3n^2 + 2n) \\
& = n^3 + 2n^2 + n + 1
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
If we're solving it using a set of linear equations, as you did, then there's a small short-cut we can use. If the degree of the polynomial is $d$ and the coefficient of the highest-order term is $a_d$, then the $D_d$ terms are all $a_d(d!)$. We can determine this by noting the coefficient of the highest order term in each $D_i$ level is $d - i + 1$ times that of the previous level, so these factors combine to form $d!$ times the $S$'s coefficient, i.e., $a_d$. We can also see this from the binomial transform as the highest order term comes from the $D_d$ values times ${n \choose d}$, so it'll be $a_d = \frac{D_d}{d!} \implies D_d = a_d(d!)$. Using this with your set of equations, we can see that $d = 3$, so $3! = 6$, gives $6 = 6A \implies A = 1$. Thus, we only need to use $3$ equations to get
$$5 = 1 + B + C + D \implies 4 = B + C + D \tag{4}\label{eq4A}$$
$$19 = 8 + 4B + 2C + D \implies 11 = 4B + 2C + D \tag{5}\label{eq5A}$$
$$49 = 27 + 9B + 3C + D \implies 22 = 9B + 3C + D \tag{6}\label{eq6A}$$
We can solve \eqref{eq4A}, \eqref{eq5A} and \eqref{eq6A} to get $B = 2$, $C = 1$ and $D = 1$, as you have done in your question text.
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