What two numbers come next in this sequence, and what is the rule?

1/2, 2, 9, 48, 300, __, __.

I came up with x_n = x_n-1(n²/(n-1)), where x₁ is 1/2.

I got this by looking at what you would have to multiply each term by to get the next. To get 2 from 1/2, you multiply by 4; to get 9 from two you multiply by 9/2; to get 48 from 9 you multiply by 48/9, which reduces to 16/3 .. you are multiplying each time by n²/(n-1).

I see that this is equivlent to Brian's answer, which almost certainly the intended response, since it also accounts for the first term. If you have that x_n-1 = [(n-1)!(n-1)]/2 , then mutliplying that by n²/(n-1) will yield:

x_n = (1/2)[(n-1)!(n-1)][n²/(n-1)]

  = (1/2)[(n-1)/(n-1)][(n-1)!n²]

  = (1/2)(1)[n(n-1)!n]

  = n(n!)/2

Which, given my formula, serves as a sort of useless proof by induction that Brian's answer is the same thing.

Posted by DJ on 2003-10-22 00:51:19