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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, 04:15

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 ..

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

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

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03 Jul 2013, 09:30

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

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

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09 Jul 2014, 01:44

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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, 20:26

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)

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

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30 Mar 2016, 17:56

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,

gmatclubot

Re: The sum of the first n positive perfect squares, where n is
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30 Mar 2016, 17:56

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