Question

Find the $n^{th}$ term of the series $10,23,60,169,494,...$

Answer 1

The successive orders of differences are
$13,37,109,335,....$
$24,72,216,....$
Thus the second order of differences is a geometrical progression in which the common ratio is $3;$ hence we may assume for the general term
$u_n=a{.3}^{n-1}+bn+c$.
To determine the constants a, b, c, make n equal to $1,2,3$ successively$;$ the $a+b+c=10,3a+2b+c=23,9a+3b+c=60;$
hence $a=6,b=1,c=3$.
Thus $u_n={6.3}^{n-1}+n+3={2.3}^n+n+3$.