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

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

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New post 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 >> :-D
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Re: The sum of the first n positive perfect squares, where n is  [#permalink]

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New post 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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New post 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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New post 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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New post 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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New post 06 Mar 2018, 13:57
Hi ALL,

Certain Quant questions on the GMAT require a bit of "playing around"; You can figure out the value of C (the unknown constant in the question) by TESTing Values - TEST any number of perfect squares, sum them up and you can solve for C. With that value in place, you can then follow the rest of the prompt, plug in N=15 and solve.

It's important to remember that most questions CAN be solved in more than one way, so keep an open mind as to what else you might do to answer a given question. Here, we actually have a question that can also be solved without the given formula. The process would take a bit longer, but there are a couple of pattern-matching shortcuts that would save some serious time.

First, name the first 15 positive perfect squares (this is knowledge that you should have memorized before Test Day):
1
4
9
16
25

36
49
64
81
100

121
144
169
196
225

Don't add them up just yet. Notice that the "unit's digit" of the ANSWERS are limited to 3 options (0, 4 or 6). Now, let's pair off those 15 numbers….

1 + 9 = ends in a 0
4 + 16 = ends in a 0
Notice a pattern…..
Every group of 5 values has a "1" and a "9" and a "4" and a "6", which means that we have lots of numbers that sum to a unit's digit of 0.

All that's left are the multiples of five: 25, 100 and 225….those numbers, when summed, end in a 0.

This means that the total sum must also end in a 0. Eliminate B, D and E.

Now, let's estimate, starting with the larger values and working to the smaller ones….
225 + 196 = over 400
169 + 144 = over 300
121 + 100 + 81 = over 300
Total so far = over 1,000
And we have all those other values to add in….The total MUST be greater than 1010. Eliminate A.

Final Answer:

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

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Re: The sum of the first n positive perfect squares, where n is   [#permalink] 16 Oct 2019, 07:34

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