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.

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]
17 May 2012, 12:43

Expert's post

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

86% (03:05) correct
14% (02:25) wrong based on 35 sessions

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?

Re: What is the sum of the first 15 positive perfect squares? [#permalink]
17 May 2012, 13:50

1

This post received KUDOS

Expert's post

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.

Re: The sum of the first n positive perfect squares, where n is [#permalink]
17 May 2012, 13:55

1

This post received 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?

Re: The sum of the first n positive perfect squares, where n is [#permalink]
17 May 2012, 17:34

Expert's post

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 ??

Re: The sum of the first n positive perfect squares, where n is [#permalink]
18 May 2012, 00:09

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. _________________

Re: The sum of the first n positive perfect squares, where n is [#permalink]
19 Dec 2012, 06:25

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.

gmatclubot

Re: The sum of the first n positive perfect squares, where n is
[#permalink]
19 Dec 2012, 06:25