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

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

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Updated on: 30 Nov 2009, 18:25
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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?

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

Originally posted by kirankp on 30 Nov 2009, 06:32.
Last edited by kirankp on 30 Nov 2009, 18:25, edited 1 time in total.
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Re: Sum of squares  [#permalink]

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30 Nov 2009, 22:06
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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 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: $$\frac{N(N + 1)(2N + 1)}{6}$$.

$$n=15$$, $$Sum=1240$$

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

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17 Feb 2010, 11:48
22
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IMO C

1 is a perfect square put n = 1

sum is $$1 = \frac{1}{3} + \frac{1}{6} + c$$

=> $$c = \frac{1}{2}$$, now put n= 15 and $$c = \frac{1}{2}$$ in $$\frac{n^3}{3} + cn^2 + \frac{n}{6}$$

This gives sum = 1240
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##### General Discussion
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Re: Sum of squares  [#permalink]

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30 Nov 2009, 13:48
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I didnt get the formula "n3/3 + cn2 + n/6", can you use {math function} to post , so we an read it correctly.

My approach would be to try $$1^2$$ or $$2^2$$ and find the value of c and then use n=15 to find the answer
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The sum of the first n positive perfect squares...PS  [#permalink]

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17 Feb 2010, 11:35
1
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
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Re: The sum of the first n positive perfect squares...PS  [#permalink]

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17 Feb 2010, 11:59
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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
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Re: The sum of the first n positive perfect squares...PS  [#permalink]

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17 Feb 2010, 12:04
1
AtifS wrote:
@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

sum = $$\frac{( 15*15*15)}{3} + \frac{15*15}{2}+ \frac{15}{6}$$= $$\frac{15*15*10}{2} + \frac{15*15}{2} + \frac{5}{2}$$

= $$\frac{(2250+225+5)}{2}$$
= $$\frac{2480}{2}$$
= 1240

On what basis you took c =1?
Try to analyze.
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Re: Sum of squares  [#permalink]

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30 Nov 2009, 22:48
If we simplify the formula its gives the same result, except for the constant, which they have tried to complicate things
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Re: Sum of squares  [#permalink]

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17 Feb 2010, 17:54
1240.

For n=1,
1=1/3 + c + 1/6
1=1/2 + c
=> c = 1/2

15*15*15/3 + 1/2*15*15 + 15/6=1240
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Re: The sum of the first n positive perfect squares...PS  [#permalink]

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18 Feb 2010, 02:07
gurpreetsingh wrote:
AtifS wrote:
@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

sum = $$\frac{( 15*15*15)}{3} + \frac{15*15}{2}+ \frac{15}{6}$$= $$\frac{15*15*10}{2} + \frac{15*15}{2} + \frac{5}{2}$$

= $$\frac{(2250+225+5)}{2}$$
= $$\frac{2480}{2}$$
= 1240

On what basis you took c =1?
Try to analyze.

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
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Re: Sum of squares  [#permalink]

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08 Dec 2010, 00:26
What is a perfect square and is this concept tested on the GMAT? Thank you.
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Re: Sum of squares  [#permalink]

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08 Dec 2010, 00:32
nonameee wrote:
What is a perfect square and is this concept tested on the GMAT? Thank you.

A perfect square, is just an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square.

Check Number Theory chapter of Math Book for more: math-number-theory-88376-60.html
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Re: Sum of squares  [#permalink]

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08 Dec 2010, 01:18
Thanks, Bunuel. Do we have to know the formula for the sum of perfect squares?
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Re: Sum of squares  [#permalink]

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11 Dec 2010, 05:25
Bunuel wrote:
nonameee wrote:
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?
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Re: Sum of squares  [#permalink]

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11 Dec 2010, 05:37
nonameee wrote:
Bunuel wrote:
nonameee wrote:
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: $$\frac{15^3}{3}+\frac{1}{2}*15^2+\frac{15}{6}=1,240$$.

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Re: Sum of squares  [#permalink]

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11 Dec 2010, 08:34
Great question.

Thanks for the explanation.
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Re: Sum of squares  [#permalink]

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20 Dec 2012, 21:57
I just memorized the formula: $$\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}$$ used to calculate sum of squares of n consecutive integers...

$$=\frac{15^3}{3}+\frac{15^2}{2}+\frac{15}{6}= 1240$$
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Re: The sum of the first n positive perfect squares, where n is  [#permalink]

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30 Jan 2013, 04:34
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.
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Re: The sum of the first n positive perfect squares, where n is  [#permalink]

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12 Jun 2013, 03:36
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 ..

It'll come >> 15[225/3+15c+1/6] = 15[(451+90c)/6] = 5[(451 + 90c)/2] ..

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

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12 Jun 2013, 03:40
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 ..

It'll come >> 15[225/3+15c+1/6] = 15[(451+90c)/6] = 5[(451 + 90c)/2] ..

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

Note that OA is C, not D. Check here: the-sum-of-the-first-n-positive-perfect-squares-where-n-is-90497.html#p831532

Hope it helps.
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Re: The sum of the first n positive perfect squares, where n is   [#permalink] 12 Jun 2013, 03:40

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