1/27 + 1/9 + 1/3 + ... + 81

I know the series is geometric. d = 1/3
Multiplying each by 1/3 to get the next value

do i use the sum of geometric series
a(1-r^n)/ 1-r

You are looking for the sum of the series?

r (not d usually, that is for arithmetic) = 3 not 1/3
a = 1/27
1/27 * 3 = 1/9
1/9 * 3 = 1/3 etc
then for sum of first n terms
a(1-r^n)/( 1-r )
If so, yes that is for the first n terms but watch out. If you want the sum for an infinite series this one does not converge r must be between -1 and + 1 so terms get smaller, not bigger with n
then for the infinite series you get
a (1)/(1-r)
because r^n goes to zero