M17 # 05

Question

What is  \frac{1}{2} + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 + ... + \left(\frac{1}{2}\right)^{20}  between?

*  \frac{1}{2}  and  \frac{2}{3} 
 *  \frac{2}{3}  and  \frac{3}{4} 
 *  \frac{3}{4}  and  \frac{9}{10} 
 *  \frac{9}{10}  and  \frac{10}{9} 
 *  \frac{10}{9}  and  \frac{3}{2}

This is a Geometric Progression:
 First term(a) 1/2 , Common ratio(r) 1/2 and n : 20+1
Sum of a GP: a(1-r^n)/(1-r)
= 1/2 (1- (1/2)^21) / (1-1/2)
(1- (1/2)^21) should be close to 1
hence answer : 1/2(1)/(1/2) = 1 (close to 1) 
D

shrouded1 · Answer

Lower Bounds 
First term is 1/2
If you add the first two terms, 1/2+1/4 you are above 3/4
So A & B can't be answers

Upper Bounds 
If this was an infinite series, the sum would be (1/2)/(1-1/2) = 1
So this sum has to be less than 1
So E can't be the answer

Between C & D 
You don't need to calculate the sum, just observe the pattern (half of the distance to 1 is always covered)
First term = 1/2
First two terms = 3/4
First three terms = 7/8
First four terms = 15/16
.. and so on ...

So it is easy to see that this sum will very quickly become greater than 9/10 (at the fourth term in fact)

So answer is (D)

Sachin9 · Answer

Bunuel,
Is there any way to solve this without using the formula?

Also, is GP tested on GMAT?

Bunuel · Answer

Check here: m17q5-72268.html

M17 # 05

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