Question

Find the sum \[\sum_{k=1}^{x}\frac{1}{2^k}\]

Answer 1

The answer says it's \[1-\frac{1}{2^x}\] , but I'm not sure how they get it. Is it not the formula for a finite geometric series?
a = 1, r = 1/2
so \[1 (\frac{1-\frac{1}{2^x}}{1-\frac{1}{2}})= 2-2^{1-x}?\]

Answer 2

\[
\frac 1 2 + \frac 1 {2^2} + \cdots + \frac 1 {2^x} =\\
\frac 1 2 \left ( 1+ \frac 1 {2^1} + \cdots + \frac 1 {2^{x-1}} \right)=
\frac 1 2\frac{1-\frac{1}{2^x}}{
\left(1-\frac{1}{2}\right)}=1-2^{-x}
\]
The first term is \( \frac 1 2\) and not 1.

Answer 3

oohh I see ok. My brain is dead right now I've been up for way too long :( Thanks for the help :)

Answer 4

yw