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

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Manager
Joined: 20 Jun 2012
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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, 05:15
Bunuel wrote:
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.

hahahahaha .. that was the funnies confusion .. I guess you are not much into texting that ":D" was a smiley like this >>
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Re: The sum of the first n positive perfect squares, where n is [#permalink]

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03 Jul 2013, 10:30
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

traditional way: put n = 1 and find c and then now substitute 15

but if u know the formula for sum of squares of n natural number (n) x (n+1 )x (2n+1 )/ 6
now directly keep n - 15

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

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09 Jul 2014, 02:44
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Re: The sum of the first n positive perfect squares, where n is [#permalink]

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12 May 2015, 21:26
Another thought.
Sum mentioned in question must NOT be a decimal. Here 15*15*15/3 = 1125, 15*15*C = 225*C, 15/6 = 2.5 which means 225*C must give a .5 after decimal part so that when it gets added up with 2.5 we will get a round number. Based on the answer choices C=0.5 should be a fit,so went for that and selected the answer (1125 + 112.5 + 2.5 = 1240)

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

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28 May 2015, 02:46

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

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30 Mar 2016, 18:56
to be honest..it was easier for me to list all the first 15 perfect square and add them up..
i got too messy with the formula..
first perfect squares:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225.

we can add up units, and see that the last digit must be 0. we can eliminate all but A and C.
i grouped the numbers to make the addition easier
4+196=200
1+9=10
25+225=250
169+121=290
36+64+100=200
16+144=160
49+81 = 130
200+200+130+160+10+290+250=400+290+300+250 = XX40 last 2 digits, so C,

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

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19 Aug 2017, 06:46
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: The sum of the first n positive perfect squares, where n is   [#permalink] 19 Aug 2017, 06:46

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