Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 26 Apr 2015, 22:14

### 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

# Powers

Author Message
TAGS:
Director
Joined: 10 Feb 2006
Posts: 664
Followers: 3

Kudos [?]: 119 [0], given: 0

Powers [#permalink]  17 Sep 2006, 16:14
= 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8

Can you guys solve this.
_________________

GMAT the final frontie!!!.

SVP
Joined: 05 Jul 2006
Posts: 1519
Followers: 5

Kudos [?]: 114 [0], given: 39

The best one who approached this problem is dahiya a 700 scorer

and this is the way he did it

2+2 = 2^2

2^2 + 2^2 = 2^3

2^3+2^3 = 2^4

2^4+2^4 = 2^5

....

.....
Senior Manager
Joined: 28 Aug 2006
Posts: 302
Followers: 10

Kudos [?]: 93 [0], given: 0

Re: Powers [#permalink]  17 Sep 2006, 20:00
= 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8

Can you guys solve this.

Hey this is simple if u know the formaula to calculate the sum of n terms of a Geometric Progression.
Excluding the first 2 in the series the remaining series is a GP with 8 terms and common ratio of 2.

So the formula of a GP whose first term is a with common ratio r and number of terms n is

a(r^n-1)/r-1 (if r>1)

a(1-r^n)/1-r (if r<1).

Here excluding the first 2 the remaining series is
2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8.
here a=2 r=2 and n=8
So sum is 2(2^8-1)/2-1 = 2 (256-1)= 510.

Adding the first two the sum has to be 510+2 = 512 = 2^9
_________________

Last edited by cicerone on 25 Sep 2008, 00:06, edited 1 time in total.
Director
Joined: 06 May 2006
Posts: 781
Followers: 3

Kudos [?]: 16 [0], given: 0

Re: Powers [#permalink]  17 Sep 2006, 22:02
= 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8

Can you guys solve this.

Sum of GP where a = 2, r = 2, n = 8

a*(r^n - 1)/(r - 1) = 2*(2^8 - 1)/(1) = 255*2 = 510

Then we have 2 + Sum = 2 + 510 = 512 = 2^9.
Director
Joined: 28 Dec 2005
Posts: 758
Followers: 1

Kudos [?]: 8 [0], given: 0

Re: Powers [#permalink]  17 Sep 2006, 22:08
= 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8

Can you guys solve this.

This is straight binary math....

2+4+8+16+32+64+128+256 = 510

Re: Powers   [#permalink] 17 Sep 2006, 22:08
Similar topics Replies Last post
Similar
Topics:
1 Powers 6 26 Dec 2011, 10:19
1 powers 2 26 Apr 2010, 02:16
Powers 2 06 Nov 2009, 23:18
PowerPrep 2 15 Aug 2005, 21:53
PS powers 2 05 Jun 2005, 07:50
Display posts from previous: Sort by