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

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30 Nov 2009, 07:32

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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]

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17 Feb 2010, 12:35

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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 _________________

"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so." Target=780 http://challengemba.blogspot.com Kudos??

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

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17 Feb 2010, 12:59

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@gurpreetsingh...yea I also chose C. but if we put c=1/2 then the answer would be 1195 correct me if I am wrong _________________

"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so." Target=780 http://challengemba.blogspot.com Kudos??

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

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12 Jun 2013, 04:36

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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 _________________

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 _________________

"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so." Target=780 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]

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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]

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12 Jun 2013, 04:40

Expert's post

stunn3r wrote:

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

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