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The sum of the first n positive perfect squares, where n is [#permalink]
 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? (A) 1,010 (B) 1,164 (C) 1,240 (D) 1,316 (E) 1,476
Last edited by kirankp on 30 Nov 2009, 19:25, edited 1 time in total.
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Re: Sum of squares [#permalink]
30 Nov 2009, 14:48
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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Re: Sum of squares [#permalink]
30 Nov 2009, 19:27
corrected it... and srini, that gives the right ans 
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Re: Sum of squares [#permalink]
30 Nov 2009, 23:06
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Re: Sum of squares [#permalink]
30 Nov 2009, 23:48
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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The sum of the first n positive perfect squares...PS [#permalink]
17 Feb 2010, 12:35
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]
17 Feb 2010, 12:48
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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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Re: The sum of the first n positive perfect squares...PS [#permalink]
17 Feb 2010, 12:59
@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]
17 Feb 2010, 13:04
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]
17 Feb 2010, 18:54
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]
18 Feb 2010, 03:07
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]
08 Dec 2010, 01:26
What is a perfect square and is this concept tested on the GMAT? Thank you.
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Re: Sum of squares [#permalink]
08 Dec 2010, 01:32
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Re: Sum of squares [#permalink]
08 Dec 2010, 02:18
Thanks, Bunuel. Do we have to know the formula for the sum of perfect squares?
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Re: Sum of squares [#permalink]
11 Dec 2010, 06:25
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]
11 Dec 2010, 06:37
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. Answer: C.
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Re: Sum of squares [#permalink]
11 Dec 2010, 09:34
Great question. Thanks for the explanation.
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Re: Sum of squares [#permalink]
20 Dec 2012, 22:57
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]
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.
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Re: The sum of the first n positive perfect squares, where n is
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30 Jan 2013, 05:34
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