AtifS wrote:
What is \(\frac{1}{2} + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 + ... + \left(\frac{1}{2}\right)^{20}\) between?
A: \(\frac{1}{2}\) and \(\frac{2}{3}\)
B: \(\frac{2}{3}\) and \(\frac{3}{4}\)
C: \(\frac{3}{4}\) and \(\frac{9}{10}\)
D: \(\frac{9}{10}\) and \(\frac{10}{9}\)
E: \(\frac{10}{9}\) and \(\frac{3}{2}\)
This is from
GMAT Club Tests and I didn't get the explanation. Can any expert help with the explanation of this question?
Fact 1 : Sum of infinite series (1/2) + (1/2)^2 + (1/2)^3 + ..... to infinity is 1. (Using infinite GP formula). So option E is out.
Next thing is how does this sum look like ?
1 term : 0.5
2 terms : 1/2 + 1/4 = 0.75
3 terms : 1/2 + 1/4 + 1/8 = 7/8 = 0.875
... and so on
Basically with every term, the distance to 1, is halfed.
Option A can't be right either as it is bounded by 2/3, which is lower than the first 2 terms
Option B can't be right either as it is bounded by 3/4, which is lower than the first 3 terms
Choosing between C & DThis is where it gets interesting
As I mentioned, with each term, the distance to 1 is halfed.
After 2 terms we are 1/4 away
After 3 terms we are 1/8 away
...
After 20 terms we will be 1/(2^20 ) away
This is a very small number, much smaller than 0.1 (which is the bound implied by C)
Hence, the answer must be DDirect Approach for the question
You can always use the sum of a GP formula, which will immediately give this sum to be equal to \(1-\frac{1}{2^{20}}\), which then makes it easy to pick D