(3^(2n-2))*(3^(2n-1)-1)/2.

I thought about it while driving home last night.

This idea follows from (2^(n-1))*(2^n-1) where 3 is subs. in for 2 and

2n-1 in for n. Division by 2 in the formula correctly calculates the

progressive sums for 1 + 3 + 9 + 27 + etc,...

Thanks, Bill