It is currently 17 Oct 2017, 10:28

### 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

# The sum of the first n positive perfect squares, where n is

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Moderator
Joined: 01 Sep 2010
Posts: 3355

Kudos [?]: 9041 [0], given: 1152

The sum of the first n positive perfect squares, where n is [#permalink]

### Show Tags

17 May 2012, 13:43
6
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

75% (02:08) correct 25% (02:27) wrong based on 155 sessions

### HideShow timer Statistics

The sum of the first n positive perfect squares, where n is a positive integer, is given by the formula n^3/3 + c*n^2 + n/6, where c is a constant. What is the sum of the first 15 positive perfect squares?

A. 1,010
B. 1,164
C. 1,240
D. 1,316
E. 1,476
[Reveal] Spoiler: OA

_________________

Kudos [?]: 9041 [0], given: 1152

Math Expert
Joined: 02 Sep 2009
Posts: 41873

Kudos [?]: 128564 [3], given: 12180

Re: What is the sum of the first 15 positive perfect squares? [#permalink]

### Show Tags

17 May 2012, 14:50
3
KUDOS
Expert's post
2
This post was
BOOKMARKED
The sum of the first n positive perfect squares, where n is a positive integer, is given by the formula n^3/3 + c*n^2 + n/6, where c is a constant. What is the sum of the first 15 positive perfect squares?
A. 1,010
B. 1,164
C. 1,240
D. 1,316
E. 1,476

Given that $$\frac{n^3}{3}+c*n^2+\frac{n}{6}$$ gives the sum of the first n positive perfect squares.

Now, for $$n=2$$ the formula should give 1^2+2^2=5 (you can equate it to 1^2 as well) --> $$\frac{2^3}{3}+c*2^2+\frac{2}{6}=5$$ --> $$c=\frac{1}{2}$$. So the formula is: $$\frac{n^3}{3}+\frac{1}{2}*n^2+\frac{n}{6}$$.

Substitute $$n=15$$ to get the sum of the first 15 positive perfect squares: $$\frac{15^3}{3}+\frac{1}{2}*15^2+\frac{15}{6}=1,240$$.

_________________

Kudos [?]: 128564 [3], given: 12180

Intern
Joined: 05 Apr 2012
Posts: 46

Kudos [?]: 37 [1], given: 12

Re: The sum of the first n positive perfect squares, where n is [#permalink]

### Show Tags

17 May 2012, 14:55
1
KUDOS
carcass wrote:
The sum of the first n positive perfect squares, where n is a positive integer, is given by the formula n^3/3 + c*n^2 + n/6, where c is a constant. What is the sum of the first 15 positive perfect squares?

A. 1,010
B. 1,164
C. 1,240
D. 1,316
E. 1,476

Hello Carcass
here it is goes c

The formula I have is

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

n= 15

Hence
15 x16 X31/6
1240

Hope this help

best regards

Kudos [?]: 37 [1], given: 12

Math Expert
Joined: 02 Sep 2009
Posts: 41873

Kudos [?]: 128564 [1], given: 12180

Re: The sum of the first n positive perfect squares, where n is [#permalink]

### Show Tags

18 May 2012, 01:09
1
KUDOS
Expert's post
carcass wrote:
Thanks keiraria

Infact I found the same formula (in that way is more simple) in this post sum-of-squares-90497.html.

So, I would ask to Bunuel two things:

first of all this is a bit unclear: for n=2 the formula should give 1^2+2^2=5 (you can equate it to 1^2 as well) . Please, can you elaborate ?? I would like to attack a problem with as methods as possible, because during the exam if you are stuck, can try to find a solution in another way.

Secondly, I think this formula is very useful but I didn't find it on Gmat math book why ??

Thanks Bunuel.

1. The formula in the stem (n^3/3 + c*n^2 + n/6) gives the sum of the first n positive perfect squares. Notice that the the value of constant c is unknown, so in order to find the sum of the first 15 positive perfect squares we should find its value. We know that the sum of the first two perfect square is 1^2+2^2=5. So, if we substitute n=2 in the formula it should equal to 5: 2^3/3 + c*2^2 + 2/6=5. From here we can find the value of c --> c=1/2 --> formula becomes: $$\frac{n^3}{3}+\frac{1}{2}*n^2+\frac{n}{6}$$ and now we can substitute n=15 to get the answer.

2. There is a direct formula (given in my post in the earlier thread) to get the sum of the first $$n$$ positive perfect squares: $$\frac{N(N + 1)(2N + 1)}{6}$$ --> if $$n=15$$ then $$Sum=\frac{N(N + 1)(2N + 1)}{6}=\frac{15(15 + 1)(2*15 + 1)}{6}=1240$$. If you know it that's fine but there are thousands of such kind formulas and you certainly cannot and should not memorize them all. For example this formula is not a must know for the GMAT.
_________________

Kudos [?]: 128564 [1], given: 12180

Senior Manager
Joined: 13 Aug 2012
Posts: 459

Kudos [?]: 540 [1], given: 11

Concentration: Marketing, Finance
GPA: 3.23
The sum of the first n positive perfect squares, where n is [#permalink]

### Show Tags

13 Dec 2012, 00:32
1
KUDOS
Let n = 1... Sum of 1 perfect squares is 1...

$$\frac{(1)^3}{3}+c*(1)^2+\frac{1}{6}=1$$
$$c=\frac{1}{2}$$

When n=15:

$$\frac{(15)^3}{3}+(15)^2/2+\frac{15}{6}=1240$$

_________________

Impossible is nothing to God.

Kudos [?]: 540 [1], given: 11

Moderator
Joined: 01 Sep 2010
Posts: 3355

Kudos [?]: 9041 [0], given: 1152

Re: The sum of the first n positive perfect squares, where n is [#permalink]

### Show Tags

17 May 2012, 18:34
Thanks keiraria

Infact I found the same formula (in that way is more simple) in this post sum-of-squares-90497.html.

So, I would ask to Bunuel two things:

first of all this is a bit unclear: for n=2 the formula should give 1^2+2^2=5 (you can equate it to 1^2 as well) . Please, can you elaborate ?? I would like to attack a problem with as methods as possible, because during the exam if you are stuck, can try to find a solution in another way.

Secondly, I think this formula is very useful but I didn't find it on Gmat math book why ??

Thanks Bunuel.
_________________

Kudos [?]: 9041 [0], given: 1152

Intern
Joined: 26 May 2012
Posts: 21

Kudos [?]: 14 [0], given: 7

Concentration: Marketing, Technology
GMAT Date: 09-17-2012
Re: The sum of the first n positive perfect squares, where n is [#permalink]

### Show Tags

19 Dec 2012, 07:25
First we need to find the constant 'c'. The easiest way to find this is for the sum of the first two perfect squares for 1 and 2 = 1 and 4 respectively.

Hence LHS = 1+4 and plug n=2 for RHS and simplify to get c = 1/2.

Plug values of n = 15 and c = 1/2 into the equation and simplify to get the answer 1240.

Option C.

Kudos [?]: 14 [0], given: 7

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16762

Kudos [?]: 273 [0], given: 0

Re: The sum of the first n positive perfect squares, where n is [#permalink]

### Show Tags

11 Jul 2015, 09:58
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 273 [0], given: 0

Manager
Joined: 22 Apr 2015
Posts: 50

Kudos [?]: 15 [0], given: 118

Location: United States
GMAT 1: 620 Q46 V27
GPA: 3.86
Re: The sum of the first n positive perfect squares, where n is [#permalink]

### Show Tags

16 Jul 2015, 15:12
Is there an easy way to simply the $$15^3/3$$ without doing it the multiplying then dividing?

Kudos [?]: 15 [0], given: 118

Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2676

Kudos [?]: 1722 [0], given: 792

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
The sum of the first n positive perfect squares, where n is [#permalink]

### Show Tags

16 Jul 2015, 16:15
xLUCAJx wrote:
Is there an easy way to simply the $$15^3/3$$ without doing it the multiplying then dividing?

Why do you need to calculate $$15^3/3$$, when you approach like this:

Once you calculate c=0.5, your equation for the sum of perfect squares becomes equal to $$n^3/3+n^2/2+n/6$$ ----> $$\frac{n(n+1)(2n+1)}{6}$$

Thus substitute n =15 in the above , we get---> 15(16)(31)/6 (cancel out 15*16/6 to get 5*8)----> 5*8*31 ----> 40*31----> 1240. This is a lot easier than calculating $$15^3/3$$
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html
Everything Related to Inequalities: http://gmatclub.com/forum/inequalities-made-easy-206653.html#p1582891
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Kudos [?]: 1722 [0], given: 792

Manager
Joined: 22 Apr 2015
Posts: 50

Kudos [?]: 15 [0], given: 118

Location: United States
GMAT 1: 620 Q46 V27
GPA: 3.86
Re: The sum of the first n positive perfect squares, where n is [#permalink]

### Show Tags

16 Jul 2015, 16:42
Engr2012 wrote:
xLUCAJx wrote:
Is there an easy way to simply the $$15^3/3$$ without doing it the multiplying then dividing?

Why do you need to calculate $$15^3/3$$, when you approach like this:

Once you calculate c=0.5, your equation for the sum of perfect squares becomes equal to $$n^3/3+n^2/2+n/6$$ ----> $$\frac{n(n+1)(2n+1)}{6}$$

Thus substitute n =15 in the above , we get---> 15(16)(31)/6 (cancel out 15*16/6 to get 5*8)----> 5*8*31 ----> 40*31----> 1240. This is a lot easier than calculating $$15^3/3$$

What rule allows you to do 15*16/6= 5*8?

Kudos [?]: 15 [0], given: 118

Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2676

Kudos [?]: 1722 [0], given: 792

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: The sum of the first n positive perfect squares, where n is [#permalink]

### Show Tags

16 Jul 2015, 17:20
xLUCAJx wrote:
Engr2012 wrote:
xLUCAJx wrote:
Is there an easy way to simply the $$15^3/3$$ without doing it the multiplying then dividing?

Why do you need to calculate $$15^3/3$$, when you approach like this:

Once you calculate c=0.5, your equation for the sum of perfect squares becomes equal to $$n^3/3+n^2/2+n/6$$ ----> $$\frac{n(n+1)(2n+1)}{6}$$

Thus substitute n =15 in the above , we get---> 15(16)(31)/6 (cancel out 15*16/6 to get 5*8)----> 5*8*31 ----> 40*31----> 1240. This is a lot easier than calculating $$15^3/3$$

What rule allows you to do 15*16/6= 5*8?

You can follow the process mentioned below:

15*16/6 ---> cancel factor of 3 out of 15 and 6 to get 5*16/2 ---> now cancel factor of 2 out of 16 and to get ---> 5*8/1---> 5*8

Look at the following reducing-fractions-or-portions-without-making-a-mistake-186285.html#p1423965
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html
Everything Related to Inequalities: http://gmatclub.com/forum/inequalities-made-easy-206653.html#p1582891
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Kudos [?]: 1722 [0], given: 792

Manager
Joined: 09 Jun 2015
Posts: 100

Kudos [?]: 7 [0], given: 0

The sum of the first n positive perfect squares, where n is [#permalink]

### Show Tags

18 Apr 2016, 01:19
carcass wrote:
The sum of the first n positive perfect squares, where n is a positive integer, is given by the formula n^3/3 + c*n^2 + n/6, where c is a constant. What is the sum of the first 15 positive perfect squares?

A. 1,010
B. 1,164
C. 1,240
D. 1,316
E. 1,476

Instead you can use this formula, which is easier; n(n+)(2*n+1)/6
15*16*31/6=40*31=1240

Kudos [?]: 7 [0], given: 0

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16762

Kudos [?]: 273 [0], given: 0

Re: The sum of the first n positive perfect squares, where n is [#permalink]

### Show Tags

15 Jul 2017, 05:41
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 273 [0], given: 0

Re: The sum of the first n positive perfect squares, where n is   [#permalink] 15 Jul 2017, 05:41
Display posts from previous: Sort by

# The sum of the first n positive perfect squares, where n is

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 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®.