trigonometry class problem: write the recursion formula for the series a(n) = 1/2(n)?

Question

obviously the parenthesis are not in this problem. but i know of no other way to represent a with the n in lower case slightly below the n and the 1 over 2 with the lower case n to the upper right of the 2.

Answer 1

So this would be read: a sub n equals one over 2 to the n power?

The first terms are 1/2, 1/4, 1/8, etc.

So: a sub 1 = 1/2

a sub (n+1) = 1/2 times a sub n

Answer 2

I don't know how to do the subscription but you usually use the carrot to denote power of.
a sub n = (1/2)^n

So the sequence would go something like this.
1/2, 1/4, 1/8, 1/16 ....

So a recursive function would look something like.
f(n) = (1/2)f(n - 1)

Answer 3

im in trig right now too, but we havent done that yet i dont think. good luck