What is the sum of n terms of the series 1/2 + 3/4 + 7/8 + : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 20 Jan 2017, 11:34

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# What is the sum of n terms of the series 1/2 + 3/4 + 7/8 +

Author Message
Senior Manager
Joined: 30 Aug 2003
Posts: 329
Location: BACARDIVILLE
Followers: 1

Kudos [?]: 18 [0], given: 0

What is the sum of n terms of the series 1/2 + 3/4 + 7/8 + [#permalink]

### Show Tags

06 Mar 2004, 15:51
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

What is the sum of n terms of the series 1/2 + 3/4 + 7/8 + 15/16 + ...equal to?

a. (2^n)-n-1
b. (2^-n)+n-1
c. (2^n)-1
d. (2^2n)-1
e.
_________________

Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

Senior Manager
Joined: 30 Aug 2003
Posts: 329
Location: BACARDIVILLE
Followers: 1

Kudos [?]: 18 [0], given: 0

### Show Tags

06 Mar 2004, 16:10

reasoning pls
_________________

Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

Senior Manager
Joined: 06 Dec 2003
Posts: 366
Location: India
Followers: 1

Kudos [?]: 11 [0], given: 0

### Show Tags

07 Mar 2004, 07:16
Quote:
What is the sum of n terms of the series 1/2 + 3/4 + 7/8 + 15/16 + ...equal to?

a. (2^n)-n-1
b. (2^-n)+n-1
c. (2^n)-1
d. (2^2n)-1
e.

Series can be written as:
(2^n - 1) / (2^n) where n runs from 1 to n
Now sum of this series is:

=sum of(1-1/(2^n))
=sum of 1 - sum of (1/(2^n)),
where sum of 1 gives n and
Now, 1/(2^n) is geometric series with a=1/2 and r=1/2
gives sumation as,
a(1-r^n)/(1-r) = (1/2)(1-(1/2)^n)/(1-1/2) = 1 - 2^(-n)
n - (1 - 2 ^ (-n))
= 2^-n + n -1

That is B

Hope, i am clear enough

dharmin
07 Mar 2004, 07:16
Display posts from previous: Sort by