Doubt in finding general term of the given sequence The following image has both the problem and its solution. I have a doubt in the solution, the details of which I have included below the image.

(Assume the terms in given series are generated from a polynomial) 
Here, the author has assumed $T_n$ as an arbitrary cubic equation (Step indicated by the RED box). 
My doubt is, why has he assumed it as a cubic equation and not a quadratic or biquadratic equation or any other degree equation? 
Please do not use Newton's forward interpolation rule as I don't know such rules. Please explain in simple terms and properties.
Kindly clarify my doubt.
 A: The author is assuming the original terms of the series are values from a polynomial at increasing integral values. Thus, when a sequence of differences causes all of the terms to be the same at some level $m$, the sequence $T_n$ represents a polynomial of degree $m$. For more information, see Theorem $1$ in Difference Tables of Sequences. In your case, taking the sequence of the third order differences based on those second order differences causes all of the values to be $6$. This means the original sequence represents a cubic equation.
Also, although the author doesn't use this, and the linked page only hints at it, the sequence of constant values are $a_n(n!)$ where $a_n$ is the coefficient of the $x^n$ term. Thus, in this case, $6 = a_3(3!) \implies a_3 = 1$. The author could have used this to determine that $a = 1$ more directly.
A: (Assuming the terms in given series are generated from a polynomial)  
When you take the difference, you reduce the degree by $1$.  
As an example, for a quadratic $T_n = n^2$, the first order differences would be in A.P: $$(n+1)^2 - n^2 = 2n+1$$
And for a cubic $T_n = n^3$, the first order differences gives you a quadratic:
$$(n+1)^3-n^3=3n^2+3n+1$$
Now if you take the difference on above quadratic, you get a sequence in A.P.
