<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] $2+12+36+80+150\dots = (1+1) + (4 + 8) + (9 + 27)+(16+64)+(25+125)+...=\sum (k^2+k^3)$
$$\sum _{k=1}^n \left(k^3+k^2\right)=\sum _{k=1}^n k^3+\sum _{k=1}^n k^2=\frac{1}{4} n^2 (n+1)^2+\frac{1}{6} n (n+1) (2 n+1)=$$
$$=\frac{1}{12} n (n+1) (n+2) (3 n+1)$$
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