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?
 A: $2+12+36+80+150\dots = (1+1) + (4 + 8) + (9 + 27)+(16+64)+(25+125)+...=\sum (k^2+k^3)$
$$\sum _{k=1}^n \left(k^3+k^2\right)=\sum _{k=1}^n k^3+\sum _{k=1}^n k^2=\frac{1}{4} n^2 (n+1)^2+\frac{1}{6} n (n+1) (2 n+1)=$$
$$=\frac{1}{12} n (n+1) (n+2) (3 n+1)$$
A: 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 $A_n=2+\sum\limits_{i=0}^{n-1}(10+11n+3n^2)=2+10n+11\dfrac{(n-1)n}2+3\dfrac{(n-1)n(2n-1)}6$
$=n^3+4n^2+5n+2.$
A: Even for an $n^{th}$ order difference series one can use the following
For a second order series, one can write: $$T_2(n)=c_3(n-2)(n-1)+c_2(n-1) +c_1$$
similarly for a third order series, one can write: $$T_3(n)=c_4(n-3)(n-2)(n-1)+c_3(n-2)(n-1) +c_2(n-1)+c_1$$
And for a $j^{th}$ order series:
$$T_j(n)=c_{j+1}(n-j)...c_4(n-3)(n-2)(n-1)+c_3(n-2)(n-1) +c_2(n-1)+c_1$$
Plug in values for $T(n)$ for different n and repetetively subtract each equation (Gaussian Elimination) to obtain the polynomial.
After finding $T(n)$ you can simply use the known results (or derive) the summations of $$\sum_{i=1}^n {i^j}$$ and plug in to get $S(n)$.
A: After @J.W. Tanner table of differences one can se that $\Delta_0=2, \Delta_1=10, \Delta_2=14, \Delta_3=6, \Delta_4=0$, then Newton's formula gives:
$$T_k= \sum_{j=0} \Delta_j {k-1 \choose j}, k\ge 1$$
$$\implies T_n=2{k-1 \choose 0}+10{k-1 \choose 1}+14 {k-1 \choose 2}+6 {k-1\choose 3}+0{k-1 \choose 4}= k^2+k^3$$
Then one can sum as
$$S_n=\sum_{k=1}^{n} [k^2+k^3]=\frac{n(n+1))(2n+1)}{6}-\left(\frac{n(n+1)}{2}\right)^2=\frac{n(n+1)(3n^2+7n+2)}{12}$$
