Work out an nth term formula for the triangular numbers 1, 3, 6, 10, 15, 21,... Please use decimals where necessary...Thankss?

The Nth triangle number is given by N(N+1)/2.

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The first number is 1.
The first difference is 2.
The second differences are all 1.
The polynomial describing this series can be written as
  1 + (n-1)/1(2 + (n-2)/2(1))
  = 1 + (n-1)((4 + n - 2)/2)
  = 1 + (n-1)(n+2)/2
  = (2 + n^2 - n + 2n - 2)/2
  = (n^2 + n)/2
  = n(n+1)/2