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

Actually, if I read your notation right, since a(n)=1/2*n, a(0)=0.

What do you have to do to 1/2(n-1) to get 1/2(n)? This will tell you the recursion formula.

Answer 2

5

Answer 3

If you're formula meant a(n) = 1 / 2^n, then the recursive formula is:

a(0) = 1
a(n) = a(n-1) / 2

Answer 4

a(n) = 1/2^n

a(n+1) = 1/2^(n+1)

a(n+1) = (1/2)*1/2^n)

Here is the recursive formula:

a(n+1)=(1/2)*a(n)

The initial value

a(0) = 1