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

It is currently 04 Sep 2015, 13:04
GMAT Club Tests

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Find the sum of the first 20 terms of the series: 1^2 + 2^2

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Manager
Manager
User avatar
Joined: 25 Jul 2006
Posts: 97
Followers: 1

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

Find the sum of the first 20 terms of the series: 1^2 + 2^2 [#permalink] New post 04 Jul 2007, 16:25
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Find the sum of the first 20 terms of the series:
1^2 + 2^2 + 3^2 + ... + 20^2

Does anyone know of a formula? Thanks.
Manager
Manager
avatar
Joined: 11 Mar 2007
Posts: 69
Followers: 1

Kudos [?]: 2 [0], given: 0

 [#permalink] New post 04 Jul 2007, 16:59
I'm pretty sure it's:

(n*(n+1)*(2n+1))/6

So if we had use 4, it would be (4*5*9)/6 = 180/6 = 30

1 + 4 + 9 + 16 = 30

20 would be:

(20*21*41)/6 = 2870

You may want to check that...
Manager
Manager
User avatar
Joined: 25 Jul 2006
Posts: 97
Followers: 1

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

 [#permalink] New post 04 Jul 2007, 17:29
mavery wrote:
I'm pretty sure it's:

(n*(n+1)*(2n+1))/6

So if we had use 4, it would be (4*5*9)/6 = 180/6 = 30

1 + 4 + 9 + 16 = 30

20 would be:

(20*21*41)/6 = 2870

You may want to check that...


wonderful - thanks!
seems to work for some no.s i tried. can you plz point me to some definitive reference on how this formula is arrived at - couldn't find it online.

also, is there a variation of this formula that is based on first term, difference, etc. (like arithmetic series Sum formula).

For example, is there a way to solve?

1. 1^2 + 3^2 + 5^2 + ...19^2

2. 4^2 + 7^2 + 10^2 + ...25^2
Manager
Manager
avatar
Joined: 11 Mar 2007
Posts: 69
Followers: 1

Kudos [?]: 2 [0], given: 0

 [#permalink] New post 05 Jul 2007, 21:04
I found it in some notes I found online...I've attached the file...

As for the other question...

I've ran across this problem that you may be able to tweak to meet your needs...

1. The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

Sol: First term a=2, common difference d=2 since even number

therefore sum to first n numbers of Arithmetic progression would be

n/2(2a+(n-1)d)

= n/2(2*2+(n-1)*2)=n(n+1) and this is equal to 79*80

therefore n=79 which is odd...
Attachments

General Math 1.doc [120 KiB]
Downloaded 64 times

To download please login or register as a user

Senior Manager
Senior Manager
avatar
Joined: 04 Jun 2007
Posts: 346
Followers: 1

Kudos [?]: 19 [0], given: 0

 [#permalink] New post 05 Jul 2007, 21:11
mavery wrote:
I found it in some notes I found online...I've attached the file...

As for the other question...

I've ran across this problem that you may be able to tweak to meet your needs...

1. The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

Sol: First term a=2, common difference d=2 since even number

therefore sum to first n numbers of Arithmetic progression would be

n/2(2a+(n-1)d)

= n/2(2*2+(n-1)*2)=n(n+1) and this is equal to 79*80

therefore n=79 which is odd...


n should really be 159 !
79 is just the number of terms in the sequence !!

And thanks for the doc. :-D
Manager
Manager
User avatar
Joined: 25 Jul 2006
Posts: 97
Followers: 1

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

 [#permalink] New post 06 Jul 2007, 15:59
1. thx for the doc - it has a formula for the sum of cubes series, which also came up for me - useful!

2. the example given is a simple linear AP (not involving squares or cubes) - not tweakable for sq or cube series (yes, agree - the ans is 159)
regardless, thanks!
  [#permalink] 06 Jul 2007, 15:59
Display posts from previous: Sort by

Find the sum of the first 20 terms of the series: 1^2 + 2^2

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.