BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Exponents

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 869
Joined: Wed Aug 26, 2009 3:49 pm
Location: California
Thanked: 13 times
Followed by:3 members

Exponents

by heshamelaziry » Sat Nov 14, 2009 10:29 pm
3 + 3 + 3 + 2 * 3^2 + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =

[spoiler]OA 3^8. I understand the first 3 terms will be 3^2. but can't go afterward ?[/spoiler]
Top
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

Re: Exponents

by Brent@GMATPrepNow » Sun Nov 15, 2009 8:11 am
heshamelaziry wrote:3 + 3 + 3 + 2 * 3^2 + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =

[spoiler]OA 3^8. I understand the first 3 terms will be 3^2. but can't go afterward ?[/spoiler]
(3 + 3 + 3) + 2 * 3^2 + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3^2 + 2 * 3^2) + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3*3^2) + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3^3) + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
3*3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3^4 + 2 * 3^4) + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3*3^4) + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3^5) + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
3^6 + 2 * 3^6 + 2 * 3^7 =
etc
3^8
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Legendary Member
Posts: 869
Joined: Wed Aug 26, 2009 3:49 pm
Location: California
Thanked: 13 times
Followed by:3 members

Re: Exponents

by heshamelaziry » Sun Nov 15, 2009 12:19 pm
Brent Hanneson wrote:
heshamelaziry wrote:3 + 3 + 3 + 2 * 3^2 + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =

[spoiler]OA 3^8. I understand the first 3 terms will be 3^2. but can't go afterward ?[/spoiler]
(3 + 3 + 3) + 2 * 3^2 + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3^2 + 2 * 3^2) + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3*3^2) + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3^3) + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
3*3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3^4 + 2 * 3^4) + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3*3^4) + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
(3^5) + 2 * 3^5 + 2 * 3^6 + 2 * 3^7 =
3^6 + 2 * 3^6 + 2 * 3^7 =
etc
3^8

How did (3^2 + 2 * 3^2) become
(3*3^2) ?
Top
Master | Next Rank: 500 Posts
Posts: 392
Joined: Thu Jan 15, 2009 12:52 pm
Location: New Jersey
Thanked: 76 times

by truplayer256 » Sun Nov 15, 2009 1:39 pm
How did (3^2 + 2 * 3^2) become
(3*3^2) ?
Factor out a 3^2 to get:

3^2(1+2)=3(3^2) or 3^3.
Top
Senior | Next Rank: 100 Posts
Posts: 98
Joined: Mon Nov 23, 2009 2:30 pm
Thanked: 26 times
Followed by:1 members

by ace_gre » Thu Dec 03, 2009 12:49 pm
Another approach to this problem. Treat the text in blue as GP..

3 + 3 + 3 + 2 * 3^2 + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7
Simplify the blue portion,
2*3^1 + 2 * 3^2 + 2 * 3^3 + 2 * 3^4 + 2 * 3^5 + 2 * 3^6 + 2 * 3^7.

Term t1=2*3^1=2*3
Common ratio(r)= t2/t1=2 * 3^2/2*3^1=3
Sum S= t1((r^n) - 1)/(r-1)
S=(2*3) * ((3^7)-1)/(3-1)
S=3*((3^7)-1)


Total series sum = 3+S=3+3*((3^7)-1) = 3(1+(3^7)-1)=3*3^7=3^8.
Top
Legendary Member
Posts: 869
Joined: Wed Aug 26, 2009 3:49 pm
Location: California
Thanked: 13 times
Followed by:3 members

by heshamelaziry » Thu Dec 03, 2009 12:56 pm
exactly what i tried to do, but failed to figure out.
Top
Post new topic Post Reply
• Page 1 of 1