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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

30 Nov 2009, 07:32

1

This post received KUDOS

23

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

70% (02:13) correct
30% (02:50) wrong based on 966 sessions

HideShow timer Statistics

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

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

The formula for counting the sum of the first \(n\) positive perfect squares is: \(\frac{N(N + 1)(2N + 1)}{6}\).

The sum of the first n positive perfect squares...PS [#permalink]

Show Tags

17 Feb 2010, 12:35

1

This post received KUDOS

The sum of the first n positive perfect squares, where n is a positive integer, is given by the formula n3/3 + cn2 + 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

Appeared on mgmat site as Problem challenge of the week.. I used the following approach to solve it but it isn't equal to any of the choices. by using the formula we may get Let's take the constant c=1 15^3/3 + 15^2 +15/6=15(15^2/3+15+1/6)=15(225/3+15+1/6)=15(75+15+1/6)=15(80+1/6)=15(481/6)=(5*481)/2=2405/2=1202.5

Any thoughts upon this question? Also don't know the OA
_________________

"I choose to rise after every fall" Target=770 http://challengemba.blogspot.com Kudos??

Thanks for clarifying and yea you are right about taking c=1, Silly me -- always loved math & have been good at it but still making silly mistakes--gotta brush up basics
_________________

"I choose to rise after every fall" Target=770 http://challengemba.blogspot.com Kudos??

Thanks, Bunuel. Do we have to know the formula for the sum of perfect squares?

No.

So the question is not a GMAT one? Am I right?

The question itself is quite realistic. You can solve it without knowing the formula I mentioned.

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: \(\frac{n^3}{3}+\frac{1}{2}*n^2+\frac{n}{6}\), (which is the same as I wrote in my first post \(\frac{n(n+1)(2n+1)}{6}\));

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

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

Show Tags

30 Jan 2013, 05:34

kirankp wrote:

The sum of the first n positive perfect squares, where n is a positive integer, is given by the formula \(\frac{n^3}{3} + c*n^2 + \frac{n}{6}\), 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

Plugging 15 into the formula and simplifying it a bit, we can get to 1125+2.5(90*C+1). The answer has to be greater than 1125. Doesn't help much as it eliminates only one answer.

2.5(90*c+1) has to fill up for whatever is left off of 1125.

Once you start looking at the options, you will notice that only 1250 works as 115 (1240-1125) can be deduced from 2.5*(90C+1) for a C value of 0.05

But trying to eliminate without actually calculating is not at all efficient, the way some of you solved it works pretty well.

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

Show Tags

12 Jun 2013, 04:36

1

This post received KUDOS

kirankp wrote:

The sum of the first n positive perfect squares, where n is a positive integer, is given by the formula \(\frac{n^3}{3} + c*n^2 + \frac{n}{6}\), 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

First of all there is a direct formula also provided above by bunuel i.e. [(n)(n+1)(2n+1)]/6

now if we do not know this and directly put 15 in place of N ..

now (450 + 90c) should be an even integer so that it should get divisible by 2, that figured out c has to be in fraction and as (450 + 90c) is an even integer answer should have "0" in the last(because it'll be multiplied by "5" outside [ ] ) .. we can eliminate B,D,E ryt away ..

for choosing between A and C. I took 1/2 as my first no. and bingo I got the answer :D
_________________

The sum of the first n positive perfect squares, where n is a positive integer, is given by the formula \(\frac{n^3}{3} + c*n^2 + \frac{n}{6}\), 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

First of all there is a direct formula also provided above by bunuel i.e. [(n)(n+1)(2n+1)]/6

now if we do not know this and directly put 15 in place of N ..

now (450 + 90c) should be an even integer so that it should get divisible by 2, that figured out c has to be in fraction and as (450 + 90c) is an even integer answer should have "0" in the last(because it'll be multiplied by "5" outside [ ] ) .. we can eliminate B,D,E ryt away ..

for choosing between A and C. I took 1/2 as my first no. and bingo I got the answer :D