Question

Find the general term and the sum of n terms of the series $12,40,90,168,280,432,.....$

Answer 1

The successive orders of difference are $28,50,78,112,152,....$
$22,28,34,40,....$
$6,6,6,....$
$0,0,...$
Hence the $n^{th}$ term $=12+28(n-1)+\frac{22(n-1)(n-2)}{2}+\frac{6(n-1)(n-2)(n-3)}{3}$
$=n^3+5n^2+6n$.
The sum of n terms may now be found by writing down the value of $\sum n^3+5\sum n^2+6\sum n$. Or we may use the formula of the present article and obtain $S_n=12n+\frac{28n(n-1)}{2}+\frac{22n(n-1)(n-2)}{3}+\frac{6n(n-1)(n-2)(n-3)}{4}$
$=\frac{n}{12}(3n^2+26n+69n+46),$
$=\frac{1}{12}n(n+1)(3n^2+23n+46)$.