Here's the way I did it. Not with a difference table. Let the nth term be an. I noticed first that the last digits go 8,3,8,3,8,3,8. Since the last digits differ by ±5, I noticed that if we subtract 3 from each one, the last digits of the new sequence will go 5,0,5,0,5,0 and therefore all be multiples of 5. Let's subtract 3 from each, and we get a sequence which has nth term5, 40, 135, 320, 625, 1080, 1715 Let's divide them by 5 and get a sequence with nth term 1, 8, 27, 64, 125, 216, 343 Aha. that is the sequence of cubes which has nth term . So we have the equation Edwin