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Take the following series D.

2
5
6
12

I then need to make a new column(and make it a matrix) to subtract the values of the previous like so

2, 3
5, 1
6, 6
12,

until eventually

2, 3, 2, 3
5, 1, 5
6, 6
12,

As you can see the result(r) is 3, given that the series could be of any size what formula would always return the last number. I also couldn't figure out how to use the MathJax with matrices. A one line formula would be very nice something like.
r = ??? D ???

  • 0
    Could be $a_n = 2+ 3 n -n(n-1)+\frac76n(n-1)(n-2)$, but could also be quite arbitrary.2012-10-13
  • 0
    oh sorry, i just realized n :P2012-10-13

1 Answers 1