Question

Find the $n^{th}$ term of the series $-1,-3,3,23,63,129,....$

Answer 1

The successive orders of differences are
$-2,6,20,40,66,....$
$8,14,20,26,...$
$6,6,6,....$
Thus the terms in the third order of differences are equal$;$ hence we may assume $u_n=A+Bn+C{n}^2+D^3$,
Where A, B, C, D have to be determined.
Putting $1,2,3,4$ for n in succession, we have four simultaneous equations, from which we obtain $A=3,B=-3,C=-2,D=1;$
hence the general term of the series is $3-3n-2n^2+n^3$.