<s>[INST] Method of summation for third order difference series: $2+12+36+80+150\dots$ I am unable to understand how one derived the formula for the $n$-th term $= an^3 +bn^2 + cn + d$, where the degree of the polynomial depends on the step at which we get a constant A.P. Here its at $2$nd step so degree $=2+1=3$.
But how do we derive this?
 [/INST] In this table, where $A_n$ is the $n^{th}$ term of your series,
$$
\begin{array}{c|c|c|c|c}
 \style{font-family:inherit}{{n}} & \style{font-family:inherit}{A_n}
& \style{font-family:inherit}{B_n} & \style{font-family:inherit}{C_n} & \style{font-family:inherit}{D_n}\\\hline
 0                                       & 2    & 10   & 14 &6\\\hline
 1                                       & 12  & 24 & 20&6 \\\hline
 2                                       & 36    & 44 & 26 \\\hline
 3                                       & 80 & 70 & \\\hline
4                                        & 150
\end{array},
$$
$D_n=6=C_{n+1}-C_n; $ $C_n=B_{n+1}-B_n$; and $B_n=A_{n+1}-A_n$.
Therefore, $C_n=14+\sum\limits_{i=0}^{n-1}6=14+6n$;
$B_n=10+\sum\limits_{i=0}^{n-1}(14+6i)=10+14n+6\dfrac{(n-1)n}2=10+11n+3n^2$;
and $A_n=2+\sum\limits_{i=0}^{n-1}(10+11n+3n^2)=2+10n+11\dfrac{(n-1)n}2+3\dfrac{(n-1)n(2n-1)}6$
$=n^3+4n^2+5n+2.$
</s>