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.
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?
Kindly clarify my doubt.
They really should have continued one step and computed the third differences: $$ 6, 6, 6, \ldots $$ (which is what they try to convey with the short and somewhat cryptic "clearly it's an A.P." comment).
A sequence that has constant third differences is given by a third degree polynomial.
(Of course there's no guarantee that the third differences will continue to be constant into the "$\ldots$" part of the sequence, but it is quite arguably the best one can do with the limited knowledge here).
By Newton's Forward Difference Formula, if $\Delta^k a_n=0$ for $k\geq 4$, that is the fourth differences are identically zero (or the third differences are constant) then
$$a_n=a_1 + \binom{n-1}{1}\Delta^1 a_1 + \binom{n-1}{2}\Delta^2 a_1 + \binom{n-1}{3}\Delta^3 a_3 $$
which is a third degree polynomial. In your case, the third differences are
$6,6,6,\dots$, therefore we have
$$a_n=1 + \binom{n-1}{1}4 + \binom{n-1}{2}10 + \binom{n-1}{3}6=1-n^2+n^3.$$
Short answer: We know that the $k$th differences of any polynomial sequence of order $k$ is constant. It follows that if the $k$th differences of a certain given sequence are constant, then the sequence must be a polynomial sequence of order $k.$
Well, to see why the property in question is true, note that given any sequence we may think of it as a sequence of partial sums of another sequence, which we can find by taking the first differences of the given sequence. On the other hand from a given sequence we may form a sequence of differences and note that if we perform these two operations consecutively on the same sequence, then they are (almost) inverse processes, just like the continuous processes of differentiation and integration.
Thus, if we understand the successive differences of sequences defined by polynomials, we shall soon see that the reverse (taking partial sums) essentially undoes the differentiation. If you now investigate this in general, you will find that it is so.
I'll only do this for quadratic sequences, but the same idea can be generalised. Let there be a sequence defined by $$an^2+bn+c,$$ then the $n+1$th terms are given by $$a(n+1)^2+b(n+1)+c,$$ and therefore we have the differences to be $$\Delta=a[(n+1)^2-n^2]+b[(n+1)-n]=a(n+1-1)(n+1+1)+b=a(2n+1)+b,$$ which is a linear sequence, which you should already know has constant differences. It therefore follows that if we perform the opposite operation of summing a linear sequence, we get a quadratic.
You may do this with a polynomial of arbitrary degree, and then the result follows by induction. In fact, it suffices to prove this only for the $k$th degree polynomial $n^k.$ And the first difference s are given by $$\Delta = (n+1)^k-n^k=(n+1-n)((n+1)^{k-1}+(n+1)^{k-2}n+\cdots+(n+1)n^{k-2}+n^{k-1})=(n+1)^{k-1}+(n+1)^{k-2}n+\cdots+(n+1)n^{k-2}+n^{k-1}),$$ where have used the well-known factorisation $$a^m-b^m=(a-b)(a^{m-1}+a^{m-2}b+\cdots +ab^{m-2}+b^{m+1}).$$ Then you now induct on $k.$
It is given: $$1,5,19,49,101,...,T_{n-3},T_{n-2},T_{n-1},T_n,...$$ Now we find: $$1) \quad 5-1,19-5,49-19,101-49,...,\color{red}{T_{n-2}-T_{n-3}},\color{green}{T_{n-1}-T_{n-2}},\color{blue}{T_n-T_{n-1}},...\\ 2) \quad 14-4,30-14,52-30,...,(\color{green}{T_{n-1}-T_{n-2}})-(\color{red}{T_{n-2}-T_{n-3}}),(\color{blue}{T_n-T_{n-1}})-(\color{green}{T_{n-1}-T_{n-2}}),...\\ 3) \quad 6,6,...,[(\color{blue}{T_n-T_{n-1}})-(\color{green}{T_{n-1}-T_{n-2}})]-[(\color{green}{T_{n-1}-T_{n-2}})-(\color{red}{T_{n-2}-T_{n-3}})]=6 \Rightarrow \\ T_n-3T_{n-1}+3T_{n-2}-T_{n-3}=6,T_1=1,T_2=5,T_3=19,$$ which is a recurrence relation, whose solution is: $$T_n=n^3-n^2+1.$$
