Integral Calculus Problem. Anyone want to try it?

Question

5
Sigma (2^-k - 2^-k - 1) =
k=1

Answer 1

5
∑(2^-k - 2^-(k+1)) =
1

5
∑2^-k( 1 - 2^-1)) =
1

... 5
½∑2^-k =
... 1

½ (2^-1 + 2^-2 + 2^-3 + 2^-4 + 2^-5) =

¼ (1 - 1/16) Sum the first 5 terms of GP with a = ½, r = ½ and n = 5)

= 15/64

Answer 2

You must have a mistake in your equation. The two 2^-k terms cancel out leaving -1, which is the sum independent of the number of k terms added.

Answer 3

Looks like the 2^-k - s^-k cancel out each time, so I'd just add -1 5 times, and get -5.