GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 22 Jun 2018, 11:51

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.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

how do we find the sum of n terms of series : (1^2) + (1^2

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Intern
Intern
avatar
Joined: 11 Oct 2008
Posts: 10
how do we find the sum of n terms of series : (1^2) + (1^2 [#permalink]

Show Tags

New post Updated on: 30 Oct 2008, 09:35
how do we find the sum of n terms of series : (1^2) + (1^2 +2^2) + (1^2 +2^2+3^2)+(1^2 +2^2+3^2+4^2) + ......

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

Originally posted by earthwork on 29 Oct 2008, 21:01.
Last edited by earthwork on 30 Oct 2008, 09:35, edited 2 times in total.
SVP
SVP
User avatar
Joined: 29 Aug 2007
Posts: 2427
Re: sequence [#permalink]

Show Tags

New post 29 Oct 2008, 22:15
earthwork wrote:
how do we find the sum of n terms of series : (1^2) + (1^2 +2^2) + (1^2 +2^2+3^2)+(1^2 +2^2+3^2+4^2) + ......


no clue.. :roll:
Probably big shots like walker have the solution.
_________________

Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html
Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html
Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html


GT

Intern
Intern
avatar
Joined: 11 Oct 2008
Posts: 10
Re: sequence [#permalink]

Show Tags

New post 29 Oct 2008, 22:40
i cud only reach till
(1^2) + (1^2 +2^2) + (1^2 +2^2+3^2)+(1^2 +2^2+3^2+4^2) + ...... = (n*1^2)+((n-1)*2^2)+((n-2)*3^2)+......
dnt even know if this is correct step!
VP
VP
avatar
Joined: 17 Jun 2008
Posts: 1479
Re: sequence [#permalink]

Show Tags

New post 29 Oct 2008, 22:52
Is it not an arithmatic progression where each subsequent term is more than the previous term by n^2 and hence, d = n^2, a = 1^2 and the sum should be a + (n-1)d = 1 + (n-1)*n^2.

I may be wrong.
Current Student
avatar
Joined: 28 Dec 2004
Posts: 3292
Location: New York City
Schools: Wharton'11 HBS'12
Re: sequence [#permalink]

Show Tags

New post 29 Oct 2008, 23:03
it cant be progressive, I think geometric series is more like it..
Intern
Intern
avatar
Joined: 30 Oct 2008
Posts: 30
Re: sequence [#permalink]

Show Tags

New post 30 Oct 2008, 06:58
The sequence can be reduced to
(n*1) + (n-1)2^2 + (n-2)3^2 + (n-3)4^2.....
Intern
Intern
avatar
Joined: 11 Oct 2008
Posts: 10
Re: sequence [#permalink]

Show Tags

New post Updated on: 30 Oct 2008, 09:46
Finally someone did help me solve it in another forum, solution below :
SOLUTION :
nth term = n(n+1)(2n+1)/6
= (2n^3 + 3n^2 + n)/6
Sum = 1/6{ 2.∑(n^3) + 3.∑(n^2) + ∑(n)}
= (1/6)[{2.(n^2)(n+1)^2/4} + 3. n(n+1)(2n+1)/6 + n(n+1)/2]
Now we can simplify this to get final answer.

Answer: (n*((n+1)^2)*(n+2)) /12

Originally posted by earthwork on 30 Oct 2008, 09:31.
Last edited by earthwork on 30 Oct 2008, 09:46, edited 2 times in total.
SVP
SVP
User avatar
Joined: 29 Aug 2007
Posts: 2427
Re: sequence [#permalink]

Show Tags

New post 30 Oct 2008, 09:37
earthwork wrote:
Finally someone did help me solve it in another forum, solution below :
SOLUTION :
nth term = n(n+1)(2n+1)/6
= (2n^3 + 3n^2 + n)/6
Sum = 1/6{ 2.∑(n^3) + 3.∑(n^2) + ∑(n)}
= (1/6)[{2.(n^2)(n+1)^2/4} + 3. n(n+1)(2n+1)/6 + n(n+1)/2]
Now we can simplify this to get final answer.


can you explain this: = (1/6)[{2.(n^2)(n+1)^2/4} + 3. n(n+1)(2n+1)/6 + n(n+1)/2]
_________________

Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html
Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html
Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html


GT

Intern
Intern
avatar
Joined: 11 Oct 2008
Posts: 10
Re: sequence [#permalink]

Show Tags

New post 30 Oct 2008, 09:45
this is a formula for sum of n terms for series:
∑k=n(n+1)/2
∑k^2=n(n+1)2n+1)/6
∑k^3=(n(n+1))^2/4

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: sequence   [#permalink] 30 Oct 2008, 09:45
Display posts from previous: Sort by

how do we find the sum of n terms of series : (1^2) + (1^2

  post reply Question banks Downloads My Bookmarks Reviews Important topics  

Moderator: chetan2u



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

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

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®.